June 2025: Paper 3, Unified chemistry

Mark scheme

Examiner's comment

Most candidates answered protons and neutrons correctly, but very few worked out the electrons for the mass spectrometry particles correctly, with 35 being the most common incorrect response. The information in the question needs to be read very carefully. With 2 marks allocated, there is a clue that the question might not be as straightforward as it first seems. Question 1 (a) (i) and Question 1 (a) (ii) proved to be excellent discriminators between students at different grades.

Notes from the mark scheme and report

• 35 protons
• 44 and 46 neutrons
• 34 electrons (1+ ions in the mass spectrometer)

Mark scheme

Examiner's comment

Most candidates gave vague responses in terms of the properties of isotopes. Some candidates were able to work out that the 158 and 162 peaks resulted from two 79Br and two 81Br isotopes respectively. The peak at 160 proved to be more difficult with many claiming that it resulted from two 80Br isotopes rather than one 79Br and one 81Br isotope. Predicting the relative abundance of the peaks was difficult, the most common response being that the peaks would have the same heights, presumably because the 79Br and 81Br isotopes have similar abundances.

Notes from the mark scheme and report

• 158 = 79Br79Br, 160 = 79Br81Br, 162 = 81Br81Br: m/z must be linked to isotopes
• 1 : 2 : 1
• IGNORE charges

Skills: M0.2 Use ratios, fractions and percentages

Mark scheme

Examiner's comment

Most candidates identified the number of peaks, for which one mark was available. The splitting patterns gave a variety of answers, with the triplet and pentet splitting from BrCH2CH2CH2Br proving to be the easier response. Most candidates predicted that BrCH2CH2Br would produce a triplet, rather than a singlet. This part of the question assessed whether candidates could identify protons with the same environment, and that they do not split one another.

Notes from the mark scheme and report

• BrCH2CH2Br: 1 peak AND singlet (equivalent protons do not split)
• BrCH2CH2CH2Br: 2 peaks
• triplet AND quintet
• Many candidates do not consider chemical environments when predicting splitting patterns.
• Protons with the same environment do not split one another.

Mark scheme

Examiner's comment

Most candidates were able to correctly draw the skeletal formula. Some candidates labelled the CH3 groups and OH connectivity was sometimes shown incorrectly as –HO or OH–. Other incorrect responses showed a main carbon chain with more, or less, than 6 carbon atoms or the CH3 side chains connected to the wrong carbon atoms. Good advice is to count the number of carbon atoms, with ‘hexan’ signifying 6 for this name.

Notes from the mark scheme and report

• must be skeletal (no CH3)
• check chain is 6 C
• DO NOT ALLOW OH–

Mark scheme

Examiner's comment

Most candidates correctly calculated the number of H atoms as 7.04  1023. Some omitted to use 18 or 130 but the mark scheme allowed such omissions so that candidates could get partial credit, provided that clear working had been shown. A common error was to show the answer to more than three significant figures despite the requirement in the question. Overall, this was one of the more confidently answered questions on the paper.

Notes from the mark scheme and report

• × 18 H per molecule
• 7.04 × 10^23, 3 SF and standard form required
• common errors: 3.91 × 10^22 (no × 18), 2.17 × 10^21 (÷ 18)

Skills: M0.1 Recognise and use expressions in decimal and ordinary form; 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

This question was answered well, with most candidates identifying that ibuprofen contains a phenol group or that it does not contain a carboxylic acid. A named functional group was required in the answer. Some described phenol incorrectly as not being an acid and this was not credited. It was rare for incorrect functional groups to be seen in candidate responses.

Notes from the mark scheme and report

• it is a phenol / has no COOH: a correct functional group must be named
• IGNORE 'too weak an acid'
• DO NOT ALLOW 'not acidic', ketone or amine

Mark scheme

Examiner's comment

There were a variety of responses to this question, with many obtaining correct answers of 206 mg for ibuprofen and 494 mg for paracetamol. From the 2:1 stoichiometry in the provided equation, the amount of ibuprofen is 4.00  10–3 mol. Candidates then needed to multiply this value by the molar mass of ibuprofen (206 g mol–1). Most candidates obtained 2.00  10–3 mol for CO2. Common subsequent mistakes resulted from candidates ignoring the 2:1 stoichiometry, working out the molar mass incorrectly, not dividing by 4 for 1 tablet and not converting to mg. Some candidates used the ideal gas equation to work out the moles of CO2. The question clearly stated RTP and the molar volume at RTP is stated on the Data Sheet. When RTP is stated, candidates are wasting valuable time by using the ideal gas equation, which is harder and opens up calculation errors, particularly with powers of 10. Overall, the question was answered well. However, many calculations were unclearly displayed, with random numbers often sprayed across the page. This could make it very difficult to find answers, especially those containing errors.

Notes from the mark scheme and report

• 2 mol ibuprofen per mol CO2
• ÷ 4 for one tablet
• 206 mg and 494 mg
• common errors 103/597 (no × 2) and 824/1976 (no ÷ 4) score 3
• ECF throughout

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; PAG1 Moles determination

Mark scheme

Examiner's comment

This unstructured calculation required candidates to work out the concentration of Ba(OH)2, to scale the concentration for the 5 times dilution and to determine the concentration of OH– ions by multiplying by 2. Many different methods were seen for working out the moles of Ba (OH)2, but the ÷5 and 2 operations were often omitted. It was common for concentrations to be shown with the wrong number of powers of 10. Many candidates used Kw or pOH confidently to obtain a value for pH. Error carried forwards (ECF) could be applied provided that the numbers used had been generated using the numbers in the question. A good number of candidates determined the pH correctly to 2 decimal places as 13.04. The commonest incorrect answer was 12.74, the result of not using 2 to find [OH–]. As with Q2b(ii), responses were often unclearly laid out and it could be difficult to find answers.

Notes from the mark scheme and report

• g dm–3 to mol dm–3 (÷ 171.3)
• dilution ÷ 5
• no pH mark without Kw or pOH
• 13.04 to 2 DP
• common errors 12.74 (no × 2), 13.74 (no ÷ 5)

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

This question was answered confidently with most candidates identified the conditions for a maximum equilibrium yield from the equation: a high pressure from fewer gaseous moles in the product and a low temperature from the exothermic reaction. This was all that the question asked but candidates often wrote far too much, considering operational conditions, rate, safely and expense. Unfortunately, candidates could end up contradicting themselves and wasting valuable time. Candidates are advised to read what the question is asking and to limit their response to this. The calculation was answered well, with most identifying that 2 significant figures is appropriate as the lowest number of significant figures present in the data provided in the question. Thus 0.48 mol dm–3 was commonly seen, rather than 0.476. The Kc expression was usually correct and most candidates rearranged this correctly to make [H2] the subject. The square root calculation showed that some candidates did not appear to have the required calculator skills. A significant number of candidates showed the Kc expression upside down and the mark scheme allowed ECF to be applied following this error.

Notes from the mark scheme and report

• high pressure because fewer gaseous moles on right (DO NOT ALLOW gaseous atoms)
• low temperature because forward reaction exothermic
• square brackets required in Kc
• square root needed
• 0.48 to 2 SF (0.476 loses a mark)
• ECF only from inverted Kc or no square root

Skills: 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; M2.4 Solve algebraic equations

Mark scheme, with the levels and indicative points

Examiner's comment

Most candidates were able to gain some marks for this Level of Response (LoR) question, with very few obtaining no marks. To achieve Level 1, candidates needed to identify at least 2 of the 3 orders with no further correct analysis. For Level 2 candidates were also expected to show the correct rate equation to match their orders or to provide a reason for why the mechanism must proceed via more than one step. Most candidates reaching Level 2 wrote a correct rate equation and usually calculated the rate constant. At this level, comparatively few produced a valid analysis of the mechanism. For a Level 3 response, all three aspects of the analysis were required: correct orders, correct rate equation and rate constant, and a correct explanation of the mechanism. Many candidates were able to provide all three aspects. Candidates usually determined the 1st order with respect to propanone from the initial rates data provided. The 1st order with respect to H+ from the rate–concentration graph was often seen. The zero order with respect to bromine from the concentration–time graph caused more problems, with many candidates incorrectly choosing 1st or 2nd order. The communication strand for this LoR question was given if the candidate had linked their orders with the evidence provided. Most candidates gave the initial rates but the evidence from the graphs was often omitted and vague. Responses such as ‘because of the graph’ are inadequate. The candidate’s response also had to be structured in a comprehensive way. This strand was usually given. beyond the minimum requirements for Level 3 including the communication strand.

Notes from the mark scheme and report

• propanone first order ([ ] × 3, rate × 3)
• Br2 zero order (constant gradient on concentration–time graph)
• H+ first order (straight line through origin)
• rate = k[CH3COCH3][H+]
• k = 4.53 × 10^-4
• Br2 in overall equation but not in rate equation (and H+ in rate equation but not overall) so more than one step

Skills: M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M3.1 Translate information between graphical, numerical and algebraic forms; PAG10 Rates of reaction – initial rates

Mark scheme

Examiner's comment

Candidates answered this question well. Overall, there has been an improvement of candidates’ ability to write systematic names involving Roman numerals Most incorrect responses either didn’t use a Roman numeral or referred to sulfur oxides or sulfides. Although not a systematic name, the mark scheme did allow sulfite. The old spellings, “sulphate” and “sulphur” are still commonly seen and were allowed.

Notes from the mark scheme and report

• sulfate(IV): IV essential
• ALLOW sulfite
• DO NOT ALLOW 'oxide' in name

Mark scheme

Examiner's comment

Candidates produced a range of answers, and the question discriminated extremely well. The higher attaining candidates often gave 6 or 7 correct formulae. The flowchart contained many clues which most candidates used. Most were able to provide at least 3-4 of the required formulae and there was no consistently wrong answer. Most candidates gave H+ and Cl– as two of the three ions, and AgCl for the white precipitate with Ag+(aq). Both S and S8 were allowed for sulfur but S2 was not allowed. As SO42– ions are present, Ag2SO4 was also allowed for the precipitate with Ag+(aq). Common errors included acid formulae such as H2SO4 in place of water, an incorrect formula for BaSO4, giving BaSO3, writing SO4– for the sulfate ion, and OH– as one of the three ‘acidic’ ions. Lower attaining candidates added formulae without using information in the question. Examples include full formulae such as HCl in the box labelled ‘ions’ and CuCl42– for the yellow solid, despite Cu being absent in the flowchart.

Notes from the mark scheme and report

• all responses must be formulae: AgCl (ALLOW Ag2SO4), BaSO4, three ions from H+, Cl–, SO42– (ALLOW ClO–, HSO4–), S (ALLOW S8
• DO NOT ALLOW S2 or 'sulfur'), H2O

Skills: PAG4 Qualitative analysis of ions

Mark scheme

Examiner's comment

Candidates found all aspects of this question extremely difficult, one of the most testing’s on this examination paper. There was important information in the flowchart to aid candidates: the ‘green solution’ indicated that Cr3+ was one of the products, and the formation of a white precipitate with Ba2+(aq) confirmed the presence of SO42–(aq). The reacting ions were provided in the flowchart. Together this information provided a framework for the equation: SO32–(aq) + Cr2O72– → Cr3+ + SO42–(aq). The equation then needed to be balanced, ideally using oxidation numbers, H2O added as a product, and finally balancing. Similar problems would be an ideal way of preparing candidates for writing equations for redox reactions.

Notes from the mark scheme and report

• 3SO32– + Cr2O72– + 8H+ → 2Cr3+ + 3SO42– + 4H2O
• DO NOT ALLOW other Cr or S species (Cr2+, Cr(OH)3, [Cr(OH)6]3–)
• IGNORE state symbols

Skills: M0.2 Use ratios, fractions and percentages

Mark scheme

Examiner's comment

Candidates produced a range of formulae, mainly from not counting H or C properly or omitting one of the elements, often Ni. A significant number of candidates wrote a structural formula rather than the empirical formula asked for in the question.

Notes from the mark scheme and report

• C6H16N2NiO6: must be empirical formula
• DO NOT ALLOW Ni(C3H6NO2)2(H2O)2

Mark scheme

Examiner's comment

The structure of the amino acid was usually drawn correctly. The nickel complex ion provided problems with H2O connectivity and superfluous use of SO4. It was expected that bonds from Ni would be shown to the O atom, and that H2O was shown correctly in all orientations. So –H2O and –O2H were not allowed. Most candidates understood the importance of showing 3D bonds as wedges and the templates on the page will have helped in this respect. However, some candidates drew impossible 3D structures with bold wedges and dashed wedges shown opposite one another. Despite the overall charge being asked for in the question, a significant number of candidates omitted the charge.

Notes from the mark scheme and report

• B = CH3CH(NH2)COOH
• 2+ charge AND bonds to O of H2O
• DO NOT ALLOW –O2H

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 drew the correct trans isomer. The common incorrect response was to show the cis isomer rotated through 180º. As connectivity had been assessed already in 4(b)(ii) for the nickel complex ion, it was not penalised for NH2 here or there would have been far more candidates who would not have been given this mark.

Notes from the mark scheme and report

• cis–trans type isomer with the bidentate ligands arranged the other way round (watch for adjacent Os)
• IGNORE connectivity to NH2 and optical isomerism within the amino acid

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

This name caused problems for most candidates. Phenyl methanol was expected, and unambiguous representations were allowed including 1- phenylmethanol, phenylmethan-1-ol and 1-phenylmethan-1-ol Many candidates incorrectly based the name on methylphenol or hydroxymethylbenzene.

Notes from the mark scheme and report

• phenylmethanol
• DO NOT ALLOW hydroxymethylbenzene or methylphenol

Mark scheme

Examiner's comment

Most candidates gave either the correct mass of LiAlH4 or the correct percentage yield. Provided that working was shown, it was possible to award some marks form a correct method. This allows ECF to be applied where a candidate had produced an incorrect number early in their calculation. Common errors included: omission of the density, dividing instead of multiplying by the density. This question discriminated well and the presence of LiAlH4 in the problem introduced a harder feature to a convention percentage yield calculation. At with the unstructured calculations elsewhere on the paper, often numbers were seen randomly placed in the answer space, making it difficult to see where errors had arisen. Candidates may be penalising themselves if they do not show clear working. It was common to see the moles of LiAlH4 being confused with the moles of the alcohol in the percentage yield calculation with moles of LiAlH4. This error is shown in Exemplar 2. The candidate has calculated the moles of C6H5COOC2H5 and C6H5CH2OH correctly and has the determined the mass of LiAlH4. This has allowed 3 of the 4 marks to be given. For the percentage yield, the candidate has then used the moles of LiAlH4 instead of C6H5CH2OH. Notice that the candidate has also given their percentage yield to 4 significant figures (SF) rather than the 3 SF asked for in the question. If the correct moles of C6H5CH2OH had been used, this error would also have prevented the mark for percentage yield from being given.

Notes from the mark scheme and report

• mass of ester = volume × density (7.875 g, 0.0525 mol)
• LiAlH4 = 0.0525 × 2.5 × 37.9 = 4.97 g
• yield 77.6%
• common errors: no density (81.4%, 4.74 g), ÷ density (85.5%, 4.46 g)
• ECF each step
• In chemistry calculations, calculator values should be carried through all steps in a calculation. This is important to eliminate errors from rounding of intermediate values.
• Any rounding should take place at the very end of the calculation.

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

Mark scheme

Examiner's comment

This question was answered correctly by most candidates, the most common error being the omission of ‘pair' in the term. Lower attaining candidates often used incorrect terms such as ‘proton donor’ and ‘electron acceptor’.

Notes from the mark scheme and report

• donates an electron pair / lone pair

Mark scheme

Examiner's comment

Candidates showed a good understanding of how to position curly arrows within an unknown mechanism. Incorrect responses were mainly in the second box. Some arrows were shown from O– to C atom instead of to the bond. It was rare for additional arrows to be added. From the responses, there was clear evidence that most candidates are becoming more confident in the accurate use of curly arrows in a mechanism and the role of lone pairs. The positioning of both ends of a curly arrow were often accurate.

Notes from the mark scheme and report

• marks solely for six curly arrows (6 = 4 marks, 5 = 3, 4 = 2, 3 = 1)
• arrow from lone pair or charge of H– / O–
• arrow from a bond must go to O in C=O and in –OC2H5
• an extra arrow is a CON
• IGNORE charges and dipoles

Mark scheme

Examiner's comment

Most candidates found this question difficult with only higher attaining candidates being able to synthesise the information provided in the method and the mechanism. In 5(a)(iv), the last step in the mechanism for Step 2 gave a clue with the two products shown with a negative charge on both O atoms. The mark scheme required an understanding that an H+ ion was involved in Step 3. With the preparation involving H– ions, responses referring to a ‘hydrogen ion’ were ambiguous and imprecise. A common incorrect response referred to hydrolysis of the ester.

Notes from the mark scheme and report

• H+ protonates the O– / negative ion to form the alcohol: must include idea of H+ being gained
• IGNORE 'to form an OH group/alcohol' alone

Mark scheme

Examiner's comment

This question looked straightforward but produced a large range of responses. The best answers were clearly communicated with different symbols being used for B, H and Na electrons in the two ions, Na+ and BH4–, from the information provided in the question. Most candidates showed BH4– with 4 covalent bonds. The expectation was that candidates would draw 3 B–H bonds using x• and a one B–H bond using a different symbol to represent a Na electron, following the requirement in the question. A dative covalent bond was often shown as the fourth B–H bond, usually as ••. The charges were often omitted and there were a number of attempts to covalently bond the Na in a single covalent molecule. Many candidates drew very small diagrams, and it was often difficult to decipher the symbols being used to represent electrons. Candidates are well-advised to adhere to the requirements and information given in the question. clearly. The candidate has used dots for B electrons, triangles for H electrons and a cross for Na electrons. Notice that this candidate has shown the full second shell of Na electrons using the same cross symbol of in the BH4– ion. Na+ was allowed showing either the full inner shell or an empty outer shell.

Notes from the mark scheme and report

• ionic Na+ and BH4–: no marks for covalent NaBH4
• 8 electrons around B in 4 B–H bonds, one bond containing the extra electron shown as Na's symbol
• charges required
• brackets and circles not required

Mark scheme

Examiner's comment

This question was answered well. Incorrect answers mainly resulted from the wrong sign for –1.24 or for 2.04 V, by adding rather than subtracting the 2 values given.

Notes from the mark scheme and report

• 0.40 – 1.64 = –1.24 V
• minus sign required

Skills: M2.2 Change the subject of an equation

Mark scheme

Examiner's comment

Candidates found this question difficult with few being given 2 marks. 1 mark was available for showing an equation with BH4– and OH– as reactants and NaBO2 and H2O as products. The equation often included O2 as a reactant and 4e–. Only higher attaining candidates were able to combine the equation successfully. This question was a very good discriminator.

Notes from the mark scheme and report

• NaBH4 + 8OH– → NaBO2 + 6H2O + 8e–
• 1 mark if correct but reversed
• species common to both sides must be cancelled

Skills: M0.2 Use ratios, fractions and percentages

Mark scheme, with the levels and indicative points

Examiner's comment

As with the other Level of Response (LoR) question (Question 3(a)), most candidates were able to gain some marks for this question, with very few obtaining no marks. Overall fewer candidates were given Level 3 for Q6 than for Q3(a). To achieve Level 1, candidates needed to calculate a value for the energy change q from mc∆T using at least two of m, c and ∆T. Candidates needed to first process the raw experimental results provided to extract the required values for the calculation. In the mc∆T calculation, the most common incorrect values used were 1.97 (the mass of Na2CO3) for m, 8.314 (the gas constant) for c and 277 (Kelvin) for ∆T. Only the lowest attaining candidates were unable to carry out this task. For Level 2, candidates were expected to determine the energy change per mole using the correct mc∆T value. Most candidates reached Level 2. A Level 3 response proved to be more elusive, requiring candidates to use an energy cycle to determine an enthalpy change. This task was far more testing than the requirements for Level 2. Some leeway was allowed for values used within the response. For m, a value in the range of 30 – 33.5 was allowed to take into account either HNO3(aq) or the mass of solution. Responses calculated per mole of Na2CO3 or HNO3 were allowed. This allowed for the fact that, even though the question described an excess of HNO3 being used in the reaction, the volume and concentration specified (30 cm3 of 1.00 dm3 HNO3) is short of being an excess. The approach taken is fully outlined in the mark scheme for this question and ensured no students were disadvantaged. The communication strand for this LoR question was given for responses that were arranged in a cohesive way. Unfortunately, a significant number of candidates produced responses that were difficult to follow with unlabelled quantities and numbers spread randomly across the page.

Notes from the mark scheme and report

• m = 33.50 g (mass of final solution, not 30 cm3)
• q = 33.50 × 4.18 × 4.0 = 560 J
• n(Na2CO3) = 1.97/106 (or via n(HNO3) = 0.0300 if acid limiting)
• ∆H2 = –30.1 (temperature rise so exothermic)
• ∆H1 = 2∆H3 – ∆H2 = +69.7 kJ mol–1
• energy cycle and signs

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; M2.4 Solve algebraic equations; PAG3 Enthalpy determination