A Level Chemistry 9701 Paper 43, May/June 2026

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9701/43 · May/June 2026

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1
a

Write the equation to represent the lattice energy, ΔHlatt, of calcium sulfide, CaS. Include state symbols.

[2]
b

Explain why the value of ΔHlatt of CaS is negative.

[1]
c

State the order of the ΔHlatt values of CaS, MgO and KCl, from most negative to least negative. Explain your answer.

[3]
d(i)

The enthalpy change of formation, ΔHf, of Cl−(g) or Mg2+(g) is the energy change when one mole of gaseous ions forms from the element in its standard state at 298 K. enthalpy change of formation of chloride ions, Cl−(g)=−243 kJ mol−1 enthalpy change of formation of magnesium ions, Mg2+(g)=+2336 kJ mol−1 The enthalpy change of formation of Cl−(g) can be calculated using two energy changes only. Identify the two energy changes.

[1]
d(ii)

The lattice energy of magnesium chloride, MgCl2, is −2493 kJ mol−1. Use this value, and other data given, to calculate the enthalpy change of formation, in kJ mol−1, of MgCl2(s).

[2]
e

Magnesium carbonate, MgCO3, decomposes when heated. MgCO3→MgO+CO2ΔS=+175 J K−1 mol−1 ΔH=+117 kJ mol−1 Predict whether the decomposition of MgCO3 is feasible at 600 ∘C. Use a calculation to explain your answer.

[3]
f

Calcium carbonate, CaCO3, also decomposes when heated. State whether the decomposition of CaCO3 occurs at a higher or lower temperature than that of MgCO3. Explain your answer.

[2]
2
a(i)

Define pH mathematically.

[1]
a(ii)

Define Kw mathematically.

[1]
b(i)

Solution A is 100 cm3 of an aqueous solution of 0.105 mol dm−3 butanoic acid, CH3CH2CH2COOH. A has pH 2.90 at 298 K. Calculate the concentration, in mol dm−3, of H+(aq) in A.

[1]
b(ii)

Calculate the value of Ka of CH3CH2CH2COOH at 298 K.

[1]
c(i)

Solution B is 100 cm3 of an aqueous solution containing 0.450 mol dm−3 CH3CH2CH2COOH. Solution C is 200 cm3 of an aqueous solution containing sodium butanoate, CH3CH2CH2COONa. B is added to C to produce 300 cm3 of solution D. The concentration of H+ ions in D is 6.31×10−5 mol dm−3. Calculate the concentration of CH3CH2CH2COONa in C. (If you were unable to calculate an answer in 2(b)(ii), then use 3.00×10−5 as the value of Ka of CH3CH2CH2COOH. This is not the correct answer.)

[3]
c(ii)

Write an equation for the reaction when H2SO4(aq) is added to D.

[1]
c(iii)

Write an equation for the reaction when KOH(aq) is added to D.

[1]
d

Predict the relative acidities of 2,2-dichlorobutanoic acid, CH3CH2CCl2COOH, and butanoic acid, CH3CH2CH2COOH. Explain your answer.

[2]
3

Nitrogen monoxide reacts with hydrogen to produce nitrogen and water. 2NO+2H2→N2+2H2O The rate equation for this reaction is shown. rate=k[NO]2[H2] Two experiments are carried out to investigate this reaction.

a

Complete Table 3.1. Table 3.1

order with respect to [NO]
order with respect to [H2]
overall order
[1]
b

In experiment 1, the rate of the reaction is 2.80×10−10 mol dm−3 s−1 when [NO] is 5.00×10−2 mol dm−3 and [H2] is 8.00×10−2 mol dm−3. Calculate the value of the rate constant, k, in experiment 1. State the units of k.

[2]
c(i)

Experiment 2 is carried out under different conditions from experiment 1. The value of k for experiment 2 is 2.13×10−5. In experiment 2, a large excess of NO is used. The initial value of [H2] is 1.20×10−3 mol dm−3. A graph of [H2] against time for experiment 2 shows that the reaction has a constant half-life. Explain why the graph shows a constant half-life, referring to: • the overall order of the reaction under these conditions • the reason why this is the overall order under these conditions.

[1]
c(ii)

Calculate the half-life, in s, of the reaction in experiment 2 and state the time taken, in s, for the value of [H2] to fall to 7.50×10−5 mol dm−3.

[2]
d(i)

The overall reaction of nitrogen monoxide with hydrogen has a three-step mechanism. 2NO+2H2→N2+2H2O Part of this mechanism is shown.

step 12NO→N2O2
step 2……………………………………⋯→………………………………………
step 3N2O+H2→N2+H2O

Complete the mechanism by writing an equation for step 2.

[2]
d(ii)

It can be deduced that step 1 is not the rate-determining step of the reaction. Explain why.

[1]
4

Some electrode potentials are given in Table 4.1. Table 4.1

electrode reactionE⊖/V
Cl2+2e−⇌2Cl−+1.36
Cu2++2e−⇌Cu+0.34
Fe(OH)3+e−⇌Fe(OH)2+OH−−0.56
Fe3++e−⇌Fe2++0.77
V2++2e−⇌V−1.13
V3++e−⇌V2+−0.26
Figure 1 of 1 for question 4
a

Complete the equation that can be used to calculate the standard Gibbs free energy change, ΔG⊖, for a cell reaction using the standard cell potential, Ecell⊖. ΔG⊖=

[1]
b

An electrochemical cell is set up using a V3+/V2+ electrode and a Cu2+/Cu electrode, both under standard conditions. Write an equation for the cell reaction that occurs and calculate the value for Ecell⊖.

[2]
c

Identify two atoms or ions from Table 4.1 that will reduce Fe3+ to Fe2+ under neutral conditions, but will not reduce Fe(OH)3 to Fe(OH)2 under alkaline conditions.

[2]
d

Identify the strongest oxidising agent and the strongest reducing agent in Table 4.1.

[2]
e

A dilute solution of NaOH(aq) is electrolysed using inert electrodes. A steady current of 2.40 A is passed for 2.00 hours. Calculate the mass, in g, of oxygen that is released at the cathode.

[3]
5
a

Cobalt is a transition element. It forms a wide range of coloured compounds that contain Co2+ ions or Co3+ ions. Explain why a transition element, such as cobalt, has variable oxidation states in its compounds.

[1]
b

Complete the electronic configurations of a Co atom and a Co2+ ion.

Co atom:[Ar]………………………………………………………………………
Co2+ ion:[Ar]………………………………………………………………………
[1]
c

Explain why compounds containing Co2+ ions are coloured.

[3]
d

A few drops of water are added to a piece of dry cobalt(II) chloride paper. Describe the colour change that is seen. Write an equation for the reaction that causes this colour change.

[2]
e(i)

Copper is a transition element. A piece of copper is added to HNO3(aq). All the copper forms Cu(NO3)2(aq). An excess of KI(aq) is added to the Cu(NO3)2(aq) in a conical flask. All the copper present forms CuI. The equation for the reaction is shown. reaction 12Cu(NO3)2+4I−→2CuI+I2+4NO3− The I2 formed in this reaction is titrated with 0.200 mol dm−3 Na2S2O3(aq). The end-point of the titration is reached when 37.60 cm3 of Na2S2O3(aq) has been added. The equation for the reaction is shown. reaction 22S2O32−+I2→S4O62−+2I− Describe two observations that can be made about the contents of the conical flask after KI(aq) has been added to Cu(NO3)2(aq).

[1]
e(ii)

Name a suitable indicator for this titration and describe the colour change at the end-point of the titration.

[1]
e(iii)

Name the type of reaction that occurs in both reactions 1 and 2.

[1]
e(iv)

Calculate the mass, in g, of the piece of copper added to HNO3(aq).

[3]
6

Transition elements form complexes with species called ligands. The ethanedioate ion, C2O42−, is a bidentate ligand.

a

Explain what is meant by bidentate ligand.

[2]
b(i)

One Fe3+ ion forms an octahedral complex with three C2O42− ions. State the formula and charge of this complex.

[1]
b(ii)

This complex shows stereoisomerism. Draw three-dimensional diagrams to show all the stereoisomers of this complex. The C2O42− ion can be represented using O⌢O.

[2]
b(iii)

State the type or types of stereoisomerism shown by this complex.

[1]
c

Hg2+ ions and iodide ions, I−, form a complex with the formula [HgI4]2−. Write the expression for Kstab of [HgI4]2−. State its units.

[2]
7
Figure 1 of 5 for question 7
Figure 2 of 5 for question 7
Figure 3 of 5 for question 7
Figure 4 of 5 for question 7
Figure 5 of 5 for question 7
a(i)

Phenol, C6H6O, can be made from benzene, C6H6, in a four-step process, as shown in Figure 7.1. C6H6→step 1C6H5NO2→step 2C6H7N→step 3C6H5N2Cl→step 4C6H6O Figure 7.1 State the reagents and conditions for step 1.

[1]
a(ii)

State the reagents and conditions for step 3.

[1]
a(iii)

Name the compound C6H5N2Cl formed in step 3.

[1]
b(i)

Phenol reacts with propanone in the presence of an acid catalyst to form compound G, as shown in Figure 7.2. Complete Table 7.1 to show the number of sp, sp2 and sp3 hybridised carbon atoms in propanone and in G. Table 7.1

spsp2sp3
propanone
G
[1]
b(ii)

State the number of peaks in the carbon-13 NMR spectrum of G.

[1]
b(iii)

The first stage of the mechanism for the reaction in Figure 7.2 involves the formation of ion J. Complete the mechanism in Figure 7.3 to show the formation of J by adding dipoles and curly arrows, as appropriate.

[1]
b(iv)

The next stage of the mechanism involves the reaction between J and phenol to form intermediate K. A proton is then lost from K to produce G. Complete Figure 7.4 to show these stages of the mechanism. Include relevant curly arrows and charges.

[3]
b(v)

Benzene is mixed with propanone in the presence of H+(aq). Predict whether benzene reacts with propanone more readily, as readily or less readily than phenol does under the same conditions. Explain your answer.

[2]
c(i)

G reacts with Br2(aq) in the dark. The organic product of this reaction is compound H. Complete the equation for this reaction in Figure 7.5.

[2]
c(ii)

Name the mechanism for this reaction.

[1]
8

Secondary amine M can be made by the three-step synthesis shown in Figure 8.1.

Figure 1 of 2 for question 8
Figure 2 of 2 for question 8
a(i)

Write an equation for the reaction of CH3COOH with SOCl2 in step 1.

[1]
a(ii)

State an alternative reagent that reacts with CH3COOH to produce CH3COCl in step 1.

[1]
a(iii)

Compound L is an amide. Draw the structure of L in Figure 8.1.

[1]
a(iv)

Draw the structure of the reagent added to CH3COCl in step 2.

[1]
a(v)

Name the mechanism of the reaction forming L in step 2.

[1]
a(vi)

Suggest a reagent for step 3.

[1]
b(i)

The proton (1H) NMR spectrum of M dissolved in D2O is obtained. Four of the proton environments, r, s, t and u, are labelled on the structure of M in Figure 8.2. Complete Table 8.1 to identify the splitting patterns for proton environments r, s, t and u. Table 8.1

proton environmentname of splitting pattern
r
s
t
u
[2]
b(ii)

The proton (1H) NMR spectrum of M dissolved in CDCl3 is obtained. State and explain how this spectrum differs from the proton (1H) NMR spectrum of M dissolved in D2O.

[2]
9

An amino acid is a compound that has both an amine group and a carboxyl group.

Figure 1 of 4 for question 9
Figure 2 of 4 for question 9
Figure 3 of 4 for question 9
Figure 4 of 4 for question 9
a

The structures of three amino acids, P, Q and R, are shown in Figure 9.1.

PQR
NH2CH(CH3)COOHNH2C(CH3)CHCOOHNH2CH2COOH

Figure 9.1 Two of the amino acids in Figure 9.1 show stereoisomerism. Draw the two stereoisomers of each of these amino acids in Figure 9.2. Name the type of stereoisomerism shown for each pair of stereoisomers. Use three-dimensional bonds where appropriate.

[3]
b(i)

The structures of amino acids T and U and their isoelectric points are shown in Table 9.1. Explain what is meant by isoelectric point.

[1]
b(ii)

Complete Table 9.2 to show the predominant species formed by amino acids T and U at the pH values shown. Table 9.2

pH 4.0pH 5.5pH 8.0
T
U
[3]
c(i)

Amino acids P, NH2CH(CH3)COOH, and R, NH2CH2COOH, can form tripeptide W, C8H15N3O4. The structure of tripeptide W can be represented as P–R–P. Draw the structure of W. The peptide functional groups formed should be displayed.

[2]
c(ii)

W reacts under suitable conditions to form two molecules of P and one molecule of R. Name the type of reaction.

[1]
c(iii)

Amino acids P and R can also form a number of polymers. One of these polymers, X, has a structure that can be represented as —[ P–R ]—n. Name the type of polymerisation involved in the formation of X.

[1]

Mark scheme

9701/43 · May/June 2026

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1a · 2 marks

  • M1

    all species

  • M2

    all state symbols

1b · 1 mark

  • 1 mark

    (ionic) bonds are formed / the ions bond

1c · 3 marks

  • M1

    MgO CaS KCl

  • M2

    CaS (and MgO) the ions / particles are more highly charged than KCl or KCl has ions with the smallest charge ora

  • M3

    MgO the ions/particles are smaller than CaS or MgO has the smallest ions ora

1d(i) · 1 mark

  • 1 mark

    (enthalpy of) atomisation (of Cl) or bond energy (Cl–Cl ) and (first) electron affinity (of chlorine)

1d(ii) · 2 marks

  • M1

    chlorine value doubled / –486 seen / 2x – 243 seen

  • M2

    answer –643 2336 +(2x – 243) –2493 = –643

1e · 3 marks

  • M1

    Correct G = H – TS and use of T = 873

  • M2

    G = –36 / –35.8 / –35.78 / –35.775 ecf T only

  • M3

    it is (feasible) and G is negative / G is less than zero UNC (Note: G now in bold) M3 is UNConditional on the basis it is a correct answer on the answer line. If the calculation has not been worked through to an answer, or has given a negative value for G, there is no M3 for ‘not feasible because G is positive’.

1f · 2 marks

  • 1 mark

    higher

  • 1 mark

    Ca2+ larger (than Mg2+) ora

  • 0 marks

    (In CaCO3) less distortion / polarisation of anion / COx y– ora all standalone ora for MgCO3 but must be stated

2a(i) · 1 mark

  • 1 mark

    –log[H+]

2a(ii) · 1 mark

  • 1 mark

    [H+][OH–] / Ka×Kb

2b(i) · 1 mark

  • 1 mark

    (10–2.9) 0.0013/0.00126/1.3x10–3/1.26x10–3

2b(ii) · 1 mark

  • 1 mark

    2bi2 / 0.105 0.001262 / 0.105 ecf bi 1.5  10–5 / 1.51  10–5 / 1.53  10–5 ecf

2c(i) · 3 marks

  • M1

    [CH3CH2CH2COOH] = 0.15 (or moles acid = 0.045 mol)

  • M2

    [CH3CH2CH2COO–Na+] in solution D = 0.036 / 0.0359 (or moles salt = 0.0108 / 0.0107 mol)

  • M3

    [CH3CH2CH2COO–Na+] in solution C = 0.054 / 0.0538 / 0.0539 / 0.0540 (or original [salt] = 0.0538 / 0.0539 / 0.0540)

2c(ii) · 1 mark

  • 1 mark

    CH3CH2CH2COO–Na+ + H+ → CH3CH2CH2COOH + Na+ or CH3CH2CH2COO– + H+ → CH3CH2CH2COOH or CH3CH2CH2COO– + H2SO4 → CH3CH2CH2COOH + HSO4– or 2CH3CH2CH2COO–Na+ + H2SO4 → 2CH3CH2CH2COOH + Na2SO4

2c(iii) · 1 mark

  • 1 mark

    CH3CH2CH2COOH + OH– → CH3CH2CH2COO– + H2O or CH3CH2CH2COOH + KOH → CH3CH2CH2COOK + H2O or CH3CH2CH2COOH + K+OH– → CH3CH2CH2COO–K+ + H2O

2d · 2 marks

  • M1

    2,2-dichlorobutanoic and chlorine is electron withdrawing / negative inductive effect / electronegative

  • M2

    weakens O–H bond / stabilises anion / stabilises conjugate base / stabilises carboxylate ion

3a · 1 mark

  • 1 mark

    2 1 3

3b · 2 marks

  • M1

    1.4x10–6

  • M2

    mol–2dm6s–1

3c(i) · 1 mark

  • 1 mark

    (overall) first order and [NO] is (almost) constant

3c(ii) · 2 marks

  • M1

    33 000 / 32 500 / 32 542 / 32 540 / 32 535 / 3.25  104, etc.

  • M2

    130 000 / 130 141 / 1.3  105 / 1.32  105, etc. ecf t1/2

3d(i) · 2 marks

  • M1

    one N2O2 and one H2 on L or one N2O and one H2O on R

  • M2

    N2O2 + H2 → N2O + H2O

3d(ii) · 1 mark

  • 1 mark

    no H2 involved in step one (and H2 in rate equation) or only one species from rate equation or does not contain species in the same molar proportion as rate equation or if step 1 is RDS then order wrt [H2] = 0 or step 1 does not include all species in rate equation or rate would be second order wrt NO only or reaction would be overall second order

4a · 1 mark

  • 1 mark

    G° = –nF E°cell

4b · 2 marks

  • 1 mark

    Cu2+ + 2V2+ → Cu + 2V3+

  • 1 mark

    (+) 0.60

4c · 2 marks

  • M1

    Cu

  • M2

    V2+

4d · 2 marks

  • M1

    Cl2 / chlorine

  • M2

    V / vanadium

4e · 3 marks

  • 3 marks

    All candidates get three marks

5a · 1 mark

  • 1 mark

    similarity in energy of (3)d and (4)s (subshells / orbitals) / A same

5b · 1 mark

  • 1 mark

    3d7 4s2 / 4s2 3d7 3d7 / 3d7 4s0

5c · 3 marks

  • M1

    the d orbitals are split / become non-degenerate / reference to E within d-subshell or E between d orbitals

  • M2

    electrons are promoted / excited / jump to higher level owtte

  • M3

    absorbing (energy of) light / photon / frequency / wavelength / h and complementary colour is seen / reflected / transmitted

5d · 2 marks

  • M1

    blue to red / blue to pink

  • M2

    CoCl2 + 6H2O → [Co(H2O)6]2+ + 2Cl– or CoCl42– + 6H2O → [Co(H2O)6]2+ + 4Cl–

5e(i) · 1 mark

  • 1 mark

    (white / brown / off white) precipitate / ppt / solid and brown (solution)

5e(ii) · 1 mark

  • 1 mark

    starch and blue / black to colourless / white / off-white

5e(iii) · 1 mark

  • 1 mark

    redox

5e(iv) · 3 marks

  • 1 mark

    moles of S2O32– = 0.2  37.6 / 1000 = 7.52  10–3

  • 1 mark

    moles of I2 = 3.76  10–3 ecf and moles of Cu = 7.52  10–3 ecf

  • 1 mark

    mass of Cu = 0.48 / 0.478 / 0.4775 / 0.47752 g ecf

6a · 2 marks

  • M1

    two lone pairs / LP

  • M2

    donated / form dative bond(s) / coordinate bond(s) to (central) metal / TM / TE atom / ion

6b(i) · 1 mark

  • 1 mark

    [Fe(C2O4)3]3– or Fe(C2O4)33–

6b(ii) · 2 marks

  • M1

    One correct isomer of [Fe(C2O4)3]3– drawn with 3D and oct

  • M2

    Two optical isomers drawn with 3D and octahedral only

6b(iii) · 1 mark

  • 1 mark

    optical only

6c · 2 marks

  • M1

    [HgI42–] / ([Hg2+][I–]4)

  • M2

    mol–4 dm12 unc

7a(i) · 1 mark

  • 1 mark

    HNO3 and H2SO4 and conc once and 25° ⩽ T ⩽ 60°

7a(ii) · 1 mark

  • 1 mark

    0° ⩽ T ⩽ 10° and HNO2 or NaNO2 + HCl

7a(iii) · 1 mark

  • 1 mark

    benzenediazonium chloride / phenyl diazonium chloride

7b(i) · 1 mark

  • 1 mark

    0 1 2 0 6 3

7b(ii) · 1 mark

  • 1 mark

    6

7b(iii) · 1 mark

  • 1 mark

    dipole and curly arrow from C=O to H+

7b(iv) · 3 marks

  • M1

    CA on leftmost diagram

  • M2

    K, showing horseshoe, + charge and H (or quinoid intermediate)

  • M3

    curly arrow on C–H bond in K back into ring

7b(v) · 2 marks

  • 1 mark

    less readily and lone pair (LP) on O (of phenol) / p-orbital on O

  • 1 mark

    is delocalised into the ring / overlaps with delocalised ring / -system and activates ring / more attractive to electrophile / electron (charge) density increased / becomes more electron rich

7c(i) · 2 marks

  • M1

    structure of product (compound H)

  • M2

    equation balanced, must have HBr

7c(ii) · 1 mark

  • 1 mark

    electrophilic substitution

8a(i) · 1 mark

  • 1 mark

    CH3COOH + SOCl2 → CH3COCl + SO2 + HCl

8a(ii) · 1 mark

  • 1 mark

    PCl3 or PCl5

8a(iii) · 1 mark

  • 1 mark

    CH3CONHCH(CH3)2

8a(iv) · 1 mark

  • 1 mark

    CH3CHNH2CH3

8a(v) · 1 mark

  • 1 mark

    (nucleophilic) addition-elimination

8a(vi) · 1 mark

  • 1 mark

    LiAlH4

8b(i) · 2 marks

  • 1 mark

    r – triplet and s – quartet/quadruplet

  • 1 mark

    t – doublet and u – multiplet

8b(ii) · 2 marks

  • M1

    one more peak and reference to N-H Subject of Q is ‘in CDCl3’ so assume this if no solvent specified. See first two ALLOWs in guidance.

  • M2

    H (of NH) is exchanged with D (in D2O) / NH + D → ND + H / H (of NH) is not exchanged with D (in CDCl3) the bits in brackets for M2 are not essential but if clearly wrong it is CON

9a · 3 marks

  • M1

    One pair of diagrams of P with correct use of 3D showing mirror images of two optical isomers

  • M2

    One pair of diagrams showing correct E and Z of Q

  • M3

    Optical or enantiomerism in the right place and Geometric(al) or E / Z or A cis/trans in the right place

9b(i) · 1 mark

  • 1 mark

    the pH at which (the amino acid is) a zwitterion / no net charge / no overall charge / no total charge electrically neutral / charge is neutral

9b(ii) · 3 marks

  • 1 mark

    two structures correct

  • 1 mark

    four structures correct

  • 1 mark

    six structures correct

9c(i) · 2 marks

  • M1

    amide links correct

  • M2

    remainder is fully correct, e.g. terminal groups not trailing bonds, neutral, zwitterion, 1+ or 1–, side/R groups CH3 – H – CH3, all H atoms on -carbons

9c(ii) · 1 mark

  • 1 mark

    hydrolysis

9c(iii) · 1 mark

  • 1 mark

    condensation

Questions

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