A Level Chemistry 9701 Paper 21, May/June 2026

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9701/21 · May/June 2026

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1
Figure 1 of 3 for question 1
Figure 2 of 3 for question 1
Figure 3 of 3 for question 1
a(i)

Complete the dot-and-cross diagram in Figure 1.1 to show the arrangement of outer shell electrons in a molecule of SF6.

[1]
a(ii)

Name the shape of a molecule of SF6.

[1]
b(i)

Table 1.1 shows information about sulfur and chlorine. Table 1.1

elementelectronic configurationfirst ionisation energy /kJ mol−1
sulfur[Ne] 3s2 3p41000
chlorine[Ne] 3s2 3p51260

Name the orbital that contains the lowest energy electron in an atom of sulfur. Draw the shape of this orbital. name ................................................................................................................................. shape

[1]
b(ii)

Construct an equation to represent the first ionisation energy of chlorine.

[1]
b(iii)

Explain the difference in the values of first ionisation energy of sulfur and chlorine in Table 1.1.

[3]
c(i)

A molecule of SCl2 is polar. Predict bond angle x in Figure 1.2 using VSEPR theory. Explain your answer. bond angle x = ..............................° explanation ........................................................................................................................ ........................................................................................................................................... ...........................................................................................................................................

[3]
c(ii)

Show on Figure 1.3 all the dipoles present in the molecule of SCl2.

[1]
d(i)

SCl2 and H2O both contain instantaneous dipole–induced dipole attractions and permanent dipole attractions. Complete Table 1.2. Table 1.2

SCl2H2O
total number of electrons per molecule10
[1]
d(ii)

Identify which molecule, SCl2 or H2O, has the stronger instantaneous dipole–induced dipole attractions. Explain your answer.

[1]
d(iii)

H2O has a greater dipole moment than SCl2. Explain why.

[1]
e

Compound X contains sulfur and chlorine only. 100.0 g of X contains 47.49 g of sulfur. The relative molecular mass, Mr, of X is 135.2. Determine the molecular formula of X. Show your working. molecular formula of X = .........................................................

[3]
2

The Contact process uses sulfur to produce sulfuric acid. stage 1 Sulfur is burned in excess air. stage 2 2SO2(g)+O2(g)⇌2SO3(g)ΔH=−196 kJ mol−1 stage 3 SO3 reacts to produce H2SO4.

a

Write an overall equation for stage 3.

[1]
b(i)

The reaction conditions for stage 2 include: • using a ratio of 1 mol SO2 to 1 mol O2 • heat in the presence of V2O5. Stage 2 is repeated using a ratio of more than 1 mol SO2 to 1 mol O2. Predict the effect on the percentage yield of SO3 at equilibrium. Use Le Chatelier’s Principle to explain your answer.

[1]
b(ii)

Explain how V2O5 affects the rate of reaction.

[2]
c(i)

CaCO3 and O2 are added to the waste gases from the Contact process before they escape. This prevents the release of SO2 into the atmosphere. equation 1 2CaCO3+2SO2+O2→2CaSO4+2CO2 State one environmental consequence of the release of SO2(g) into the atmosphere.

[1]
c(ii)

Use the ideal gas equation to calculate the volume of O2(g) that reacts with 0.500 mol of SO2 at 400 ∘C and 101 kPa in equation 1. Include units in your answer. volume of O2(g) = ............................................................... units ...............................................................

[4]
3

Separate samples of NaCl, NaBr and NaI each react with concentrated H2SO4. Table 3.1 shows the observations for each reaction. Table 3.1

reactioncompoundobservations
1NaCl• effervescence • steamy fumes that turn blue litmus paper red
2NaBr• effervescence • steamy fumes initially that turn blue litmus paper red • brown vapour • pungent smell of SO2
3NaI• effervescence • steamy fumes initially that turn blue litmus paper red • purple vapour • pungent smell of SO2 and then a rotten egg smell of H2S
a(i)

Write an equation for reaction 1.

[1]
a(ii)

Identify the reactant that behaves as a Brønsted–Lowry base in reaction 1.

[1]
b

State the formulae of the species responsible for the brown vapour in reaction 2 and for the purple vapour in reaction 3. brown vapour ............................................................... purple vapour ...............................................................

[1]
c

Explain why reaction 3 produces purple vapour.

[1]
d

Complete Table 3.2. Table 3.2

speciesoxidation number of sulfur
H2SO4
SO2
H2S
[2]
e

Explain why reaction 3 produces H2S but reaction 2 does not produce H2S.

[1]
f(i)

Describe what is observed when Ag+(aq), followed by NH3(aq), are added to a sample of NaI(aq). observation after addition of Ag+(aq) ................................................................................. observation after addition of NH3(aq) ................................................................................

[2]
f(ii)

Write an ionic equation for the reaction of Ag+(aq) with NaI(aq).

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

Compound Z is oxidised to produce organic compound W, C5H6O7, by heating under reflux with an excess of acidified Cr2O72−(aq). Describe the colour change in this reaction. colour of mixture at start of the reaction ............................................................................ colour of mixture after the reaction ....................................................................................

[1]
a(ii)

The complete oxidation of 0.025 mol of Z requires 0.050 mol of Cr2O72−(aq). W is produced in this reaction. Balance the equation for this reaction. .....C5H12O4 + .....Cr2O72− + .....H+ → .....C5H6O7 + .....Cr3+ + .....H2O

[1]
b(i)

Compound A is a structural isomer of Z. Complete Table 4.1 to show what is observed when separate samples of A are added to each reagent. State ‘none’ if no reaction occurs. Table 4.1

reagentobservation with A
Na(s)
2,4-dinitrophenylhydrazine (2,4-DNPH)
Fehling’s reagentnone
[2]
b(ii)

Suggest why Fehling’s reagent does not react with A but does react with aldehyde functional groups.

[1]
b(iii)

Compound A is heated under reflux with an excess of acidified Cr2O72−(aq) to produce organic compound Y. Draw the structure of Y.

[2]
c

Compound B reacts with an excess of LiAlH4 to produce organic compound C. Complete the equation in Figure 4.3 to show the reaction of B with an excess of LiAlH4 to produce C. [H] represents one atom of hydrogen from the reducing agent, LiAlH4.

[2]
5

Figure 5.1 describes a sequence of reactions starting with ethene to produce poly(chloroethene), PVC. ethene →reaction 1 1,2-dichloroethane →reaction 2 chloroethene →reaction 3 poly(chloroethene) Figure 5.1

Figure 1 of 2 for question 5
Figure 2 of 2 for question 5
a

Complete Table 5.1. Table 5.1

compoundtype of hybridisation of carbon atoms
1,2-dichloroethane
chloroethene
[2]
b

In reaction 1, ethene reacts with Cl2(g) at room temperature. Name the mechanism for reaction 1.

[1]
c

Describe what is meant by heterolytic fission of a covalent bond.

[2]
d

Complete Figure 5.2 to show the mechanism for the reaction of ethene with Cl2 in reaction 1. Include charges, dipoles, lone pairs of electrons and curly arrows, as appropriate.

[4]
e

When 1,2-dichloroethane is heated in reaction 2, thermal decomposition occurs. The reaction is endothermic. State the type of reaction that occurs.

[1]
f

Draw the repeat unit of poly(chloroethene).

[1]
g(i)

Figure 5.3 shows the mass spectrum of chloroethene. Identify the species responsible for the peaks at m/e=64 and m/e=27. peak at m/e=64 .............................................................................................................. peak at m/e=27 ..............................................................................................................

[2]
g(ii)

Explain the difference in relative abundance of the peaks at m/e=62 and m/e=64.

[2]

Mark scheme

9701/21 · May/June 2026

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1a(i) · 1 mark

  • 1 mark

    Dot-and-cross diagram of SF6 showing 12 electrons around S and complete octets around each F

1a(ii) · 1 mark

  • 1 mark

    octahedral

1b(i) · 1 mark

  • 1 mark

    circle/sphere AND label 1s

1b(ii) · 1 mark

  • 1 mark

    Cl(g) → Cl+(g) + e(–)

1b(iii) · 3 marks

  • M1

    (small) increase in shielding in Cl is outweighed by / similar shielding

  • M2

    increased nuclear charge in Cl / Cl has greater nuclear charge

  • M3

    Cl has greater nuclear attraction

1c(i) · 3 marks

  • M1

    104.5(°)

  • M2

    2 bonding pairs (of electrons) and 2 lone pairs (of electrons)

  • M3

    lone pairs (of electrons) repel more than bonding pairs (of electrons)

1c(ii) · 1 mark

  • 1 mark

    S as δ+ AND Both Cl as δ–

1d(i) · 1 mark

  • 1 mark

    total number of electrons per molecule: SCl2 = 50, H2O = 10

1d(ii) · 1 mark

  • 1 mark

    SCl2 AND greater number of electrons (per molecule) OR H2O has weaker id (id) as it has smaller number of electrons (per molecule)

1d(iii) · 1 mark

  • 1 mark

    electronegativity difference (between H and O) is greater (than S and Cl)

1e · 3 marks

  • M1

    mass Cl = 52.51

  • M2

    47.49 / 32.1 : 52.51/35.5 (= 1.48 :1.48) (empirical formula =) SCl

  • M3

    (Mr(X) = 135.2) molecular formula) = S2Cl2

2a · 1 mark

  • 1 mark

    SO3 + H2O → H2SO4

2b(i) · 1 mark

  • 1 mark

    increases yield AND equilibrium shifts to right (hand side)

2b(ii) · 2 marks

  • M1

    it / rate (of reaction) increases

  • M2

    (it) lowers EA OR (used) to lower EA

2c(i) · 1 mark

  • 1 mark

    acid rain

2c(ii) · 4 marks

  • M1

    PV=nRT

  • M2

    number of mol of O2 required: n = 0.5 ÷ 2 = 0.25

  • M3

    temperature conversion to kelvin = 400 + 273 = 673

  • M4

    vol = (M2 ⨯ 8.31 ⨯ M3) ÷ 101 000 = 0.0138431 AND m3

3a(i) · 1 mark

  • 1 mark

    H2SO4 + NaCl → NaHSO4 + HCl.

3a(ii) · 1 mark

  • 1 mark

    Cl– / NaCl

3b · 1 mark

  • 1 mark

    Brown vapour ……Br2 Purple vapour …...I2

3c · 1 mark

  • 1 mark

    (purple vapour is produced when) HI / iodide (anions) are oxidised (by H2SO4) OR (sulfuric) acid / H2SO4 oxidises I– (to iodine)

3d · 2 marks

  • 2 marks

    H2SO4: (+)VI OR (+)6; SO2: (+)IV OR (+)4; H2S: –II OR –2 (all 3 correct for 2 marks, 2 correct for 1 mark)

3e · 1 mark

  • 1 mark

    iodide / I– is a stronger reducing agent (than bromide / Br–) OR HBr / bromide / Br− is a weaker reducing agent (than iodide / I−) OR HBr / Bromide / Br− can’t reduce S(IV) / SO2

3f(i) · 2 marks

  • M1

    (observation after addition of Ag+(aq)) ….(pale) yellow precipitate

  • M2

    (observation after addition of NH3(aq)) ….precipitate remains / precipitate does not dissolve / precipitate does not disappear

3f(ii) · 1 mark

  • 1 mark

    Ag+ + I– → AgI

4a(i) · 1 mark

  • 1 mark

    (colour of mixture at start of reaction) …………orange AND (colour of mixture after the reaction) …………green

4a(ii) · 1 mark

  • 1 mark

    C5H12O4 + 2Cr2O72- + 16H+ → C5H6O7 + 4Cr3+ + 11H2O

4b(i) · 2 marks

  • M1

    Na(s): effervescence / fizzing / bubbling

  • M2

    2,4-DNPH: none AND Fehling's reagent: none

4b(ii) · 1 mark

  • 1 mark

    Alcohols / Compound A are / is not oxidised by Fehling’s reagent (Aldehydes are oxidised by Fehling’s reagent) OR Aldehydes are oxidised by Fehling’s reagent (Alcohols / Compound A are / is not oxidised by Fehling’s reagent)

4b(iii) · 2 marks

  • 2 marks

    HOOCC=OC=OCH2COOH or fully correct displayed/skeletal structure

4c · 2 marks

  • M1

    structure of C

  • M2

    equation fully correct: reactant + 8[H] → C + H2O

5a · 2 marks

  • M1

    1,2 dichloroethane: sp3

  • M2

    chloroethene: sp2

5b · 1 mark

  • 1 mark

    electrophilic addition

5c · 2 marks

  • M1

    when a (covalent) bond / it is broken

  • M2

    one atom takes both of the electrons from the shared pair

5d · 4 marks

  • M1

    curly arrow from C=C bond to Cl closest to C=C

  • M2

    correct dipoles Cl δ+—Cl δ– AND curly arrow from bond of Cl δ+—Cl δ– to Cl δ–

  • M3

    correct carbocation intermediate

  • M4

    curly arrow from lone pair of : Cl– to C+ of intermediate

5e · 1 mark

  • 1 mark

    elimination

5f · 1 mark

  • 1 mark

    -[CH2-CHCl]- or correct repeat unit structure

5g(i) · 2 marks

  • M1

    [CH2CH37Cl ]+

  • M2

    [CH2CH]+

5g(ii) · 2 marks

  • M1

    2 chlorine isotopes (exist)

  • M2

    35Cl is 3 times more abundant than 37Cl

Questions

  1. 1
    Question 117 marks
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  2. 2
    Question 29 marks
    Mark scheme ›
  3. 3
    Question 310 marks
    Mark scheme ›
  4. 4
    Question 49 marks
    Mark scheme ›
  5. 5
    Question 515 marks
    Mark scheme ›

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