June 2022: Paper 1, Periodic table, elements and physical chemistry

Answer

D

Mark scheme

Examiner's comment

This question proved to be difficult, with only the most able candidates selecting the correct answer of D. A was often given as an incorrect answer as candidates recognised that AgCl would be the only halide precipitate to show a change with dilute ammonia but did not realise that as it would redissolve, it would be the only one not in the filtrate.

Notes from the mark scheme and report

• AgCl dissolves in dilute NH3
• AgBr and AgI do not

Skills: PAG4 Qualitative analysis of ions

Answer

C

Mark scheme

Examiner's comment

This question was quite well answered with many candidates identifying the correct response as C. Candidates had to link the volume of gas with the moles of each gas and then match to the stoichiometry of the equation. Some candidates calculated the moles of gas and then appeared to choose an answer at random.

Notes from the mark scheme and report

• gas volume ratio 2:1:2 equals mole ratio

Skills: M0.2 Use ratios, fractions and percentages

Answer

B

Mark scheme

Examiner's comment

This was a challenging question, the correct answer being B. The candidates would be advised to use the space around the question to perform the calculation. A was a common error as the Al ion was seen as the obvious choice without doing any calculations due to being Al3+.

Answer

C

Mark scheme

Examiner's comment

The correct answer C required candidates to calculate the moles of manganate (VII) ions and then to use the stoichiometry of the equation to calculate the moles of H2O2 before calculating the concentration. Option A provided a distractor with the incorrect ratio. Option B was obtained without the reference to the molar ratio.

Notes from the mark scheme and report

• 2:5 ratio MnO4–:H2O2

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

Answer

B

Mark scheme

Examiner's comment

The correct answer B required the candidate to calculate the number of moles of the gas using Avogadro’s number. Then link this numerical value to the Mr of the gas so that the formula could be deduced from calculating the Mr of the molecules suggested. Incorrect answers, often if appeared that option C had been guessed as an answer, rather than being arrived at by mistake.

Answer

B

Mark scheme

Examiner's comment

This was generally well answered. The key to candidates quickly arriving at the correct answer of B was to focus on identifying the number of CN- ions and calculating the total of 18 negatively charge ions. More successful candidates could clearly see that the combination of iron ions must add up to the total of 18 positively charged ions. Some candidates lost time here by working out formulae and trying to write out the structure of Prussian blue.

Notes from the mark scheme and report

• 18 CN– so total positive charge 18 from 7 Fe

Skills: M2.4 Solve algebraic equations

Answer

A

Mark scheme

Examiner's comment

This was another challenging question, with fewer candidates choosing the correct answer of A. The distractor, D, was obtained by swapping products and reactants in the calculation or in an incorrect enthalpy cycle. Candidates should have ruled out C and D straight away as enthalpy of combustion is always exothermic. Where working was shown, candidates omitted one or more of the multiples in the bonds formed or broken.

Answer

A

Mark scheme

Examiner's comment

This question was quite well answered with many candidates realising that for a first order reaction, determination of the rate constant, k, from the constant half-life, t1/2, they should use the relationship: k = ln 2/t1/2. A common error was option B, as many used 1/t used for ln 2/t1/2.

Notes from the mark scheme and report

• k = ln 2 / t½

Skills: 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

Answer

A

Mark scheme

Examiner's comment

This question proved to be difficult, with fewer candidates selecting the correct answer of A. Option C was the most common distractor as many candidates did not know the standard state of bromine. Option B was selected by those candidates who confused the definitions of average bond enthalpy with the enthalpy change of atomisation.

Notes from the mark scheme and report

• 1 mol of gaseous atoms from the element in its standard state (liquid Br2)

Answer

A

Mark scheme

Examiner's comment

Only some candidates chose the correct answer of A. Option D was a common incorrect answer through not recognising that the enthalpy change for condensation is negative.

Answer

D

Mark scheme

Examiner's comment

This was well answered, with many candidates understanding that the addition of an alkali will react with the H+ ions and will keep the position of equilibrium on the right hand side.

Notes from the mark scheme and report

• already yellow (A– form) so adding OH– causes no change

Answer

C

Mark scheme

Examiner's comment

This was generally well answered. The key to candidates arriving at the correct answer ,C, was to write out the equation for ammonia reacting with water. The correct identification of the acids can then be made. Option B was chosen as the candidate identified the B-L bases and option D was chosen by candidates who linked up a B-L acid – base pair.

Mark scheme

Examiner's comment

In the interests of fairness, we decided to award the mark for this question to all candidates as this question related to theory [or a process] that was not required within our specification.

Notes from the mark scheme and report

• mark scheme: award the mark regardless of response (question discounted)

Answer

A

Mark scheme

Examiner's comment

This question was not well answered, with many candidates giving the other options rather than the correct answer of A. The candidates who were successful tended to draw out the shape of the molecules.

Notes from the mark scheme and report

• CO2, SiCl4 and SF6 are all symmetrical so non-polar

Answer

B

Mark scheme

Examiner's comment

This was another challenging question, with fewer candidates choosing the correct answer of B. Successful candidates often had the electronic configuration written down and focused on the higher energy electrons as to whether they were paired or not. Missing the 4s1 in Cu was the main omission to identify the correct answer. Section B overview

Notes from the mark scheme and report

• Cu+ is 3d10

Mark scheme

Examiner's comment

Almost all candidates produced a graph which showed the correct shape curve, although they sometimes incorrectly identified the quantities on the axes, e.g. rate versus time. Those who produced an incorrect diagram were able to discuss ‘more’ particles reaching the activation energy threshold. A few limited their answer to discussing successful collisions. Candidates needed to identify which curve was the higher temperature and to make sure that the curves did intersect more than once.

Notes from the mark scheme and report

• axes: number of molecules vs energy (DO NOT ALLOW atoms/enthalpy)
• curve from origin, not touching x-axis
• higher-T curve has lower peak shifted right, crossing once
• more molecules with energy above Ea
• DO NOT ALLOW 'lowers Ea'
• IGNORE more successful collisions

Skills: M3.1 Translate information between graphical, numerical and algebraic forms

Mark scheme

Examiner's comment

Many candidates were able to explain clearly the second order dependence on [NO] but many found it more challenging to identify and explain the first order relationship for [CO], where the candidate had to consider the second order dependence on [NO] to work out the effect on rate for the change in [CO]. Most gave a rate equation of the correct format based on the orders they had stated, and worked out a value for k. Although some did not determine the correct units, there was an encouraging number who managed this either from memory or by cancelling. The most common error was not including s-1

Notes from the mark scheme and report

• NO second order from expts 1 and 2
• CO first order needs both concentrations changing (expts 2 and 3: rate × 16)
• k = 3.37 × 10^6 dm6 mol–2 s–1
• common errors 4.65 × 10^9 mol–3 dm9 s–1 (CO second order) and 2446 mol–1 dm3 s–1 (CO zero order) score 4 with ECF

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; PAG10 Rates of reaction – initial rates

Mark scheme

Examiner's comment

Many candidates recognised the need for 2NO2 in the first step, with slightly less then adding an appropriate second step. Errors included unbalanced species and unbalanced charges. A wide range of suggested possible mechanisms were seen, including several which led to multiple correct versions of the overall equation. Both balanced equations needed to combine to form the overall equation. However, a few inventive candidates gave a first step of NO2 + NO2 → 2NO2, followed by the full balanced equation for the second step – this was not given marks.

Notes from the mark scheme and report

• step 1 must have 2NO2 only on the left
• second mark dependent on the first
• steps must add up to the overall equation

Mark scheme

Examiner's comment

Many candidates made some attempt at this unusual twist on a Q = mcΔT calculation, but only a minority of candidates produced a fully correct answer. A temperature of 84.9 oC was seen much more often than 52.4 oC. The calculation of n(AgNO3) was performed well. Common errors were not dividing the moles by 2, not recognising the need to calculate the energy released per mole of AgNO3, confusing the mass of water with the mass of the silver nitrate solid or of using an energy in kJ alongside a value of c based on J.

Notes from the mark scheme and report

• n(AgNO3) = 0.0400
• energy = 678 × 0.0400/2 = 13.56 kJ (2 mol AgNO3 per ∆H)
• ∆T = 32.4
• add 20.0 gives 52.4 °C to 3 sig figs
• common errors 32.4 (∆T only) and 84.9 (no ÷2) score 3

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; PAG3 Enthalpy determination

Mark scheme

Examiner's comment

Although there were a few well explained, correct answers, this question proved challenging for most candidates. Candidates needed to link the changes as a proportion (e.g. half) to be given the mark. Candidates often only considered the change to energy or mass, but not both, e.g. dealt only with the decrease in moles reacting leading to a smaller temperature rise.

Notes from the mark scheme and report

• same maximum temperature AND half the energy/moles AND half the mass/volume

Skills: M0.2 Use ratios, fractions and percentages; PAG3 Enthalpy determination

Mark scheme

Examiner's comment

The majority of candidates gave the required definition correctly, frequently going beyond to include ideas about standard states and conditions. The smaller number of incorrect responses covered a range of errors, including assuming that 1 mole of elements were reacting and (perhaps as a result of later work on lattice enthalpies) bringing in a requirement for the compound or the elements to be gaseous.

Notes from the mark scheme and report

• 1 mole of compound formed from its elements
• DO NOT ALLOW 1 mole of element
• DO NOT ALLOW 'gaseous elements'

Mark scheme

Examiner's comment

This calculation proved difficult. Many candidates did not deduce the correct cycle. Of those who did, a significant majority did not appreciate the stoichiometry and/or use the correct signs from the cycle. Many students scored 1 mark for this question which was gained as an error carried forward, providing all the values had been used in the enthalpy cycle. -90 was a common error.

Notes from the mark scheme and report

• 4∆fH(NO) = –1172 + 1716 – 184 = +360
• ÷4 = +90
• –90 scores 1

Skills: M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

Candidates needed to explain that entropy is a measure of the dispersal of energy in a system which is greater, the more disordered a system. A good number of correct answers but quite a few did not get across the required idea of ‘a measure of’.

Notes from the mark scheme and report

• measure of dispersal of energy/disorder

Mark scheme

Examiner's comment

Almost all candidates had a good attempt at this calculation, with many gaining full marks. Most were able to calculate the entropy change. Almost all could reproduce the equation for free energy. Of those who did not get the correct final answer, the most common error was not converting the entropy value into kJ and/or the temperature to K There were also a few candidates who did not manipulate the equation correctly.

Notes from the mark scheme and report

• ∆S = +284 J K–1 mol–1 converted to 0.284 kJ
• T = 298 K
• ∆H = ∆G + T∆S = –2587
• common errors –2664.9 (25 °C), 81960 (∆S in J), –2756.6 (wrong sign of ∆S) score 2

Skills: M0.0 Recognise and make use of appropriate units in calculations; 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

Many candidates realised the significance of a negative free energy change, but they seemed to struggle to link this to the previous question, giving a more general answer about feasibility, rather than using their values of entropy change and enthalpy change to determine feasibility. If temperature of feasibility was calculated as -9110K many still thought that meant a temperature existed below this which was feasible.

Notes from the mark scheme and report

• ∆S positive AND ∆H negative
• so ∆G negative at all temperatures
• ALLOW ECF from signs in (c)(ii)

Mark scheme

Examiner's comment

Few candidates scored full marks here, with some candidates not increasing the 1st IE across the period, and many getting the dip for the wrong element (not O) and finishing too high with Ne. Less successful candidates had clearly confused it with MP trend across the period.

Notes from the mark scheme and report

• general increase from B to Ne with Ne lower than He
• O lower than N but higher than C, F higher than O

Skills: M3.1 Translate information between graphical, numerical and algebraic forms

Mark scheme

Examiner's comment

This question proved demanding for candidates, with many simply quoting a molar value taken from the graph and converting into standard form. Of those who recognised the need to use the Avogadro constant, a few tried to multiply it by the molar ionisation enthalpy. For those who worked out the correct answer, several lost marks due to the requirement of 2 significant figures.

Notes from the mark scheme and report

• about 500/6.02 × 10^23 = 8.3 × 10^–22 kJ
• must be 2 sig figs AND standard form

Skills: M0.1 Recognise and use expressions in decimal and ordinary form; M0.3 Estimate results; M1.1 Use an appropriate number of significant figures

Mark scheme

Examiner's comment

Some very wordy responses to this straightforward question were seen, with many candidates going onto the extra pages. Candidates are reminded that keeping responses concise and to the point can make their answer clearer and potentially avoid contradicting themselves in the process. There were a good number of excellent answers seen, although many candidates were distracted by arguments about the stability of full shells or subshells rather than explaining why nuclear attraction would be greater. Some students mixed up the explanation for Be/Li with the difference in ionisation energies between groups 5 and 6 (electron pair repulsion).

Notes from the mark scheme and report

• comparison needed for each mark
• He: outer electron in closer/lower shell
• Be: more protons with same shell/similar shielding
• greater nuclear attraction
• IGNORE 'effective nuclear charge', atomic number increases, just 'greater attraction'
• IGNORE 'different sub-shell'

Mark scheme

Examiner's comment

Many candidates did not achieve full marks as they did not discriminate between the three different comparisons that were being tested in Question 18 (c) and 18 (d). Few candidates did identify the electron being in the s or p but then explained the ionisation energy in terms of full subshells, electron pair repulsion and described the 2s sub-shell as closer to the nucleus rather than as lower energy.

Notes from the mark scheme and report

• Be electron removed from 2s, B from 2p
• 2p is at higher energy
• DO NOT ALLOW 'shell'
• DO NOT ALLOW unpaired electron argument
• IGNORE distance and shielding

Mark scheme

Examiner's comment

Almost all candidates calculated the titres correctly, but a significant number were penalised for recording 22.70 as 22.7. A significant number also used this result and the trial titre to derive their mean value. There were also very small numbers of candidates who used all 4, or all of titres 1, 2 and 3, to calculate their mean.

Notes from the mark scheme and report

• titres to 2 d.p. (22.70 not 22.7)
• mean from concordant 22.45 and 22.55 = 22.50
• DO NOT ALLOW use of trial
• Links to the legacy coursework tasks and PAG practice question sets can be found on OCR Interchange. Exam hints for students can be found at: https://www.ocr.org.uk/Images/592305-exam-hints-for-students.pdf.

Skills: M1.2 Find arithmetic means; PAG2 Acid–base titration

Mark scheme

Examiner's comment

Candidates made good progress with this calculation, many gaining 4 or 5 marks, including error carried forward from incorrect titres. Common errors included, in various combinations: not converting the final answer into mg, not converting volume to dm3, missed ratio, multiplying the moles in 10cm3 acid by 10 instead of 25 and/or wrong Mr. Responses to Question 19 (a) (ii) often featured rows of figures and random sums without a single word about what the figures, or sums, were set to calculate. Candidates should remember to provide written indications of what it is they are working out – presenting the calculations without any annotations can make it harder for error carried forward marks to be given if there is an error in their calculation. The exemplar here shows a good use of annotation. There is a clear indication of the mathematical process so that the error carried forward is easily identified and the candidate gains the method marks.

Notes from the mark scheme and report

• ÷2 (dibasic)
• × 25 to scale 10.0 cm3 to 250.0 cm3
• × 118
• ÷4 tablets and convert to mg = 498
• common errors 996 (no ÷2), 19.9 (no × 25)
• 503 from titre 22.725 scores 5 by ECF

Skills: M0.0 Recognise and make use of appropriate units in calculations; 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; PAG2 Acid–base titration

Mark scheme

Examiner's comment

Many candidates did not answer the question and instead described what a buffer was. Very few candidates correctly explained that a weak acid was being added to a base, sometimes mentioning the formation of the salt or conjugate base. The majority also did not include the importance of there being excess acid, or some acid remaining, after the partial neutralisation.

Notes from the mark scheme and report

• acid in excess/partially neutralised AND glycolate ions (conjugate base/salt) formed

Mark scheme

Examiner's comment

Most candidates were able to derive a value for Ka from pKa and calculate the number of moles of glycolic acid and potassium hydroxide reacting. Less were successful in determining the moles or concentrations present in the buffer solution causing many to get the common error of 3.57. Many candidates tried to calculate pH for the weak acid, without considering changes to concentrations or the buffering effect. Clarity of working is essential and in questions such as this, candidates are advised to include word descriptions of what they are calculating, even if it is abbreviations such as ‘n’ for number of moles.

Notes from the mark scheme and report

• n(HA) left = 0.0450 – 0.0250 = 0.0200
• n(A–) = 0.0250
• Ka = 10^–3.83
• pH = 3.93 (2 d.p.)
• common errors 3.57 (HA not reduced) scores 3, 3.75 (concentrations from question) scores 2

Skills: 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 correctly described the ammonia reacting with the glycolic acid or with hydrogen ions, although some thought that ammonia was acidic. Many of them then went on to say that “the equilibrium will move to the right” without realising that the equilibrium had not itself appeared within the question, and so they needed to write it out to gain marks. A few candidates thought that ammonia was an acid, due to the 3 x Hs in the molecule.

Notes from the mark scheme and report

• NH3/OH– reacts with H+ (or the acid HA)
• equilibrium equation HA ⇌ H+ + A– must be shown AND shifts right

Mark scheme, with the levels and indicative points

Examiner's comment

This Level of Response question was generally well answered with many candidates achieving maximum marks by simply considering what was required in the question. Responses were often split between a calculation on the main paper and the conditions explanation on extra pages. The calculation errors included no shift or incorrect shift in the equilibrium values. Not calculating the concentration or incorrectly multiplying by 8 rather than dividing by 8. Some candidates attempted a ‘hybrid’ calculation of Kp by trying to calculate a mole fraction and partial pressures. There was a number of candidates who confidently worked out the value of Kc. There were also some very good analyses of the operational conditions. Many of those who had done well on the calculation treated the explanation as an afterthought, not giving it enough attention to give them an answer that would access Level 3. An example of a complete answer, showing a good level of communication in the description and the calculation layout, achieving Level 3 (6 marks) is shown above. This candidate gave a clear method using an “ICE” table to calculate the number of moles at equilibrium. The calculation of the new concentrations can be seen by the use of [ ] and division by 8. The candidate then shows the Kc expression and substitutes the numerical values before successfully calculating the value and includes the units. This is followed by a concise explanation of the conditions used by industry.

Notes from the mark scheme and report

• equilibrium moles N2 1.12, H2 3.36 (0.08 and 0.24 reacted)
• divide by 8.00 dm3
• Kc = 0.0386 dm6 mol–2
• low temperature/high pressure for yield vs high temperature for rate, cost and safety of high pressure
• catalyst allows lower temperature

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

Candidates coped well with this question, but many candidates did not gain the mark due to ambiguous statements. Some identified the forward reaction as endothermic, but stated that Kp decreased which was given in the question. Others simply stated that the forward reaction was endothermic as the reverse reaction was exothermic.

Notes from the mark scheme and report

• equilibrium shifts left as T decreases AND forward reaction endothermic

Mark scheme

Examiner's comment

Many candidates gave the correct reason to agree with student 2. Those who agreed with student 1 did not see the equation as a heterogeneous equilibrium system. There were a small number of responses agreeing with student 2 but for the wrong reason – such as a confusion about how the position of equilibrium can change when the value of Kp stays constant. Candidates are advised to read through and address all parts of the question as a minority of students didn’t identify which student was correct but gave a correct explanation.

Notes from the mark scheme and report

• Student 2 correct AND same number of moles of GAS on each side (solids ignored)

Mark scheme

Examiner's comment

Very few candidates made two valid statements where both clearly indicated an idea of relative rate – in almost all cases one of the descriptions would be about quantity of gas rather than rate of gas production. Some candidates identified a precipitate being formed, colour change, or gave a general answer of the reaction happening quicker.

Notes from the mark scheme and report

• both needed for 1 mark: Ca fizzes faster AND Ca dissolves more quickly
• IGNORE 'more bubbles' (rate needed)
• IGNORE exothermic

Mark scheme

Examiner's comment

Some candidates coped well with this question which was based on the AS part of the specification and gained both marks. More candidates gained 1 mark through writing one half equation, usually the oxidation of magnesium. Common errors were for chlorine to featuring in the reduction half equation and the lack of electrons in their answers. Very few candidates mixed up the oxidation and reduction equations.

Notes from the mark scheme and report

• Mg → Mg2+ + 2e–
• 2H+ + 2e– → H2
• only species that change
• 1 mark if both correct but labelled the wrong way round

Mark scheme

Examiner's comment

Although few candidates got the correct final answer, however almost all achieved some marks from this calculation through error carried forward, with marks spread across the available range. Almost all candidates were able to find the concentrations of hydrogen and hence hydroxide ions. A few candidates successfully used p[OH-] method. Most were able to calculate the moles of hydroxide ions in 250cm3. Many then did not realise the need to half this number to find the moles of barium, and/or used the Mr for barium hydroxide instead of barium oxide.

Notes from the mark scheme and report

• ÷4 for 250 cm3
• ÷2 for Ba(OH)2
• × 153.3 (BaO) = 2.53 g
• common errors 5.05 (no ÷2) and 2.82 (Mr of Ba(OH)2) score 4, 5.65 scores 3

Skills: M0.2 Use ratios, fractions and percentages; M0.4 Use calculators to find and use power, exponential and logarithmic functions; M1.1 Use an appropriate number of significant figures; M2.5 Use logarithms in relation to quantities that range over several orders of magnitude

Mark scheme

Examiner's comment

This question was answered well, with many candidates giving one of the equations in the ‘ALLOW’ part of the mark scheme. Those candidates who did not gain this mark gave full equations or missed out state symbols.

Notes from the mark scheme and report

• Ba2+(aq) + 2H+(aq) + SO4 2–(aq) + 2OH–(aq) → BaSO4(s) + 2H2O(l)
• ALLOW H+ + OH– → H2O OR Ba2+ + SO4 2– → BaSO4 with state symbols

Mark scheme

Examiner's comment

This proved a difficult question for most candidates. Most were able to correctly calculate the moles of fertiliser by converting kg to g. The next step was to deduce that 5 moles of calcium carbonate would be required for each mole of Z and multiply by 5, rather than the common error of dividing by 5. Few candidates were able to multiply by 100/95, to account for the impurities in limestone, with many multiplying by 95/100.

Notes from the mark scheme and report

• n(Z) = 1500/1080.5
• × 5 for CaCO3
• × 100.1 (DO NOT ALLOW 100)
• ÷ 0.950 = 731 g
• common errors 146 (no × 5), 660 (× 0.95), 29.3 (÷5) score 2

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures

Mark scheme

Examiner's comment

This was another very challenging question using an unfamiliar mineral. Most candidates identified a formula of salts containing both magnesium and calcium, or carbonates of the separate elements. Only the most successful candidates were able to give the correct formula. Common errors, for those who solved the formulae, were the use of “4”HCl in balancing and the absence of state symbols.

Notes from the mark scheme and report

• Mg3Ca(CO3)4(s) + 8HCl(aq) → 3MgCl2(aq) + CaCl2(aq) + 4H2O(l) + 4CO2(g)
• balancing/state symbols mark dependent on correct formulae

Mark scheme

Examiner's comment

There were some good answers to this standard question, but candidates did not gain the mark due to missing one of the three key ideas of “donation, electron pairs and forming two coordinate bonds”.

Notes from the mark scheme and report

• donates two electron/lone pairs AND forms two coordinate bonds (to the metal ion)

Mark scheme

Examiner's comment

Most candidates did not seem to understand how to apply their knowledge of what an empirical formula should look like to this situation. Many giving the formula of the complex ion instead. Those candidates who were successful often had the incorrect overall charge on the ion with 3+ as the common error, not considering the 2 Cl- ligands. Conversely, most candidates did very well at representing the isomers of the complex. Most were able to use the wedge and broken line notation to suggest the geometry of what they were drawing. Although a small number drew the trans isomer in three different orientations, most correctly drew the trans and cis forms and managed to show the optical isomers – even when they did not make use of the top pair of answer boxes to assist them with their mirror images. Candidates are advised to focus on the mirror image and avoid subsequent rotations and inversions of the molecule. This can often lead to two of the same isomers being drawn. Candidates showed a good accuracy with their connectivity of the N atom.

Notes from the mark scheme and report

• empirical formula FeC4H16N4Cl2 with 1+ charge (DO NOT ALLOW Fe(NH2CH2CH2NH2)2Cl2)
• two cis mirror images and one trans
• correct wedges (2 out, 2 in, 2 in plane)
• Fe–N connectivity errors penalised once
• 1 mark if all three shown with Cl but no N

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

Most students scored this mark, although several gave no response.

Notes from the mark scheme and report

• Fe2+ + 2OH– → Fe(OH)2
• ALLOW complexed formulae
• IGNORE state symbols

Mark scheme

Examiner's comment

Although a spread of marks across the full available range was seen, a good proportion of candidates gained 3 or 4 marks. Most candidates were able to produce a balanced equation, but candidates should take care cancelling out any species present on both side of the equation, e.g. the hydroxide ions. A common error within the formula of iron (III) hydroxide was to place the number of hydroxide ions within the brackets, e.g. Fe(OH3). Candidates are advised to read the instructions contained within the equation and to use or comment on all the data presented. When commenting on electrode potentials, candidates should avoid the use of higher/lower as these phrases are meaningless due to the negative signs involved.

Notes from the mark scheme and report

• brown ppt is Fe(OH)3/Fe(OH)2 oxidised
• E of system 1 more negative than system 2
• system 1 shifts left (or system 2 right)
• 4Fe(OH)2 + O2 + 2H2O → 4Fe(OH)3
• IGNORE lower/higher
• DO NOT ALLOW uncancelled species

Mark scheme, with the levels and indicative points

Examiner's comment

Answers were distributed across all 3 levels of achievement. Most of the candidates managed to identify at least some of the substances. Of the equations, the reaction of copper (II) oxide with nitric acid was most regularly seen correct, although many students could also represent the ligand replacement in Experiment 1. Many candidates were able to calculate Mr for gas D but some of those suggesting SO2 as a possible formula preferred to have an equation in experiment 2 producing hydrogen. A few candidates used the Mr to suggest that the gas was 2O2 and as such candidates found the equation between copper and sulphuric acid challenging. A good number of candidates identified F and G, recognising what they had learned from their work on redox titrations, and some were able to reproduce the equation. Incorrect formula of copper (I) iodide (CuI2) was a common error. Many candidates made no attempt at identifying changes in oxidation states. Candidates are advised to address all parts of the question in order to access the higher levels and to allow sufficient time to attempt the LoR questions. The candidate has clearly set out the three experiments, written the equation and followed with a clear identification of each of the formulae B-G. The calculation is well laid out and the response clearly indicates how the final value has been calculated. Oxidation numbers have then been commented on where appropriate. The colours of the transition metal ions and complexes can be reviewed at: A Level Chemistry A - Colours of inorganic ions and complexes (ocr.org.uk) If any of your students’ results are not as expected, you may wish to consider one of our post-results services. For full information about the options available visit the OCR website. We send a weekly roundup to tell you about important updates. You can also sign up for your subject specific updates. If you haven’t already, sign up here. Attend one of our popular CPD courses to hear directly from a senior assessor or drop in to a Q&A session. Most of our courses are delivered live via an online platform, so you can attend from any location. Please find details for all our courses on the relevant subject page on our website or visit OCR professional development. ExamBuilder is the question builder platform for a range of our GCSE, A Level, Cambridge Nationals and Cambridge Technicals qualifications. Find out more. ExamBuilder is free for all OCR centres with an Interchange account and gives you unlimited users per centre. We need an Interchange username to validate the identity of your centre's first user account for ExamBuilder. If you do not have an Interchange account please contact your centre administrator (usually the Exams Officer) to request a username, or nominate an existing Interchange user in your department. Review students' exam performance with our free online results analysis tool. It is available for all GCSEs, AS and A Levels and Cambridge Nationals. It allows you to: • review and run analysis reports on exam performance • analyse results at question and/or topic level • compare your centre with OCR national averages • identify trends across the centre • facilitate effective planning and delivery of courses • identify areas of the curriculum where students excel or struggle • help pinpoint strengths and weaknesses of students and teaching departments. Find out more. Post-results services Keep up-to-date OCR Professional Development Signed up for ExamBuilder?

Notes from the mark scheme and report

• B = [CuCl4]2–
• D: n = 45/24000, M = 64 so SO2, C = CuSO4 with Cu 0 to +2 and S +6 to +4
• E = Cu(NO3)2
• F = CuI (white) and G = I2 with Cu +2 to +1 and I –1 to 0
• Level 3 needs Mr calculation, most equations and some oxidation number changes
• The AS delivery guide ‘Theme: Identifying Unknowns’ https://www.ocr.org.uk/Images/208563-identifying-unknowns.pdf ‘Transition elements’: Delivery Guide for OCR AS/A Level Chemistry A

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