A Level Chemistry 9701 Paper 22, May/June 2025

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9701/22 · May/June 2025

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1
a(i)

Diamond and graphite are both crystalline solids made from carbon atoms. Graphite conducts electricity. Diamond does not conduct electricity. Name the type of lattice structure present in the crystalline solids diamond and graphite.

[1]
a(ii)

Explain how graphite conducts electricity.

[1]
a(iii)

Explain why diamond does not conduct electricity.

[1]
b(i)

Separate samples of phosphorus(V) chloride and silicon(IV) chloride are each added to an excess of cold water. Write an equation for each reaction.

[2]
b(ii)

Describe the appearance of phosphorus(V) chloride and silicon(IV) chloride at room temperature. phosphorus(V) chloride ..................................................................................................... silicon(IV) chloride ............................................................................................................

[2]
b(iii)

Compare the appearance of the mixtures produced when each reaction is complete.

[2]
c(i)

Some oxides are amphoteric. Describe what is meant by amphoteric.

[1]
c(ii)

Identify the formula of a Period 3 oxide that is amphoteric.

[1]
d(i)

The melting points of different oxides are shown in Table 1.1. Table 1.1

oxidemelting point/∘Cforce of attraction broken during melting
SO2−73
H2O0
SiO21610
Na2O1132
MgO2852

Complete Table 1.1 by identifying the strongest force of attraction in each oxide that is broken during melting. Use the abbreviations below. i.d. = instantaneous dipole–induced dipole p.d. = permanent dipole–permanent dipole H = hydrogen bond C = covalent bond I = ionic bond

[3]
d(ii)

A student suggests the following hypothesis. The stronger the covalent bond between atoms in non-metal oxides, the higher the melting point. Use Table 1.1 to deduce if this hypothesis is true or false or if there is not enough information to make a conclusion. Explain your answer.

[2]
2

A sample of iron contains three different isotopes and has a relative atomic mass, Ar, of 55.8.

a(i)

Define relative atomic mass.

[2]
a(ii)

Table 2.1 shows the abundances of two of the isotopes present in the sample of iron. Table 2.1

isotoperelative isotopic massabundance/%
54Fe53.96.0
56Fe55.991.9

Use Table 2.1 to calculate the relative isotopic mass of the third isotope of iron in the sample. Show your working. relative isotopic mass = ..............................

[2]
b

Deduce the number of pairs of electrons in the 3d sub-shell in an iron(II) ion.

[1]
c

Sketch the shape of the lowest energy orbital in the shell with principal quantum number n=2.

[1]
d

Complete Table 2.2 to show information about particles in one ion of 56Fe3+. Table 2.2

particlenumber of particles present in one ion of 56Fe3+
electrons
30
[2]
e

The atomic radius of iron is 1.26×10−10 m. Suggest the change to the radius, if any, after an iron atom reacts to produce an Fe3+ ion. Explain your answer.

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

C3H6 reacts with HBr(g) in an addition reaction. Define addition reaction.

[1]
a(ii)

Complete Fig. 3.1 to show the mechanism for the addition reaction between C3H6 and HBr to produce 2-bromopropane. Include charges, dipoles, lone pairs of electrons and curly arrows, as appropriate. Fig. 3.1

[4]
a(iii)

Explain why the major product of this reaction is 2-bromopropane rather than 1-bromopropane.

[2]
b(i)

H2(g) and nickel are added to alkene X. Fig. 3.2 shows how the concentration of X changes with time. Fig. 3.2 Use Fig. 3.2 to identify the limiting reagent in this reaction. Explain your answer.

[1]
b(ii)

Use Fig. 3.2 to describe how the gradient changes as the reaction proceeds. State what this shows about the rate during this reaction.

[1]
c

Hydrocarbon Y contains two C=C groups. There are no other functional groups present. Y reacts with an excess of H2(g) to produce hexane, C6H14. Table 3.1

bondenergy/kJ mol−1
C−C350
C=C610
C≡C840
H−H436
C−H410

Use Table 3.1 to calculate the enthalpy change per mole of C6H14 produced in this reaction. enthalpy change = .............................. kJ mol−1

[2]
d(i)

Hydrocarbon Y reacts with H2(g). Fig. 3.3 shows the distribution of energies of H2(g) at temperature T. Area A represents the number of molecules with energy greater than or equal to the activation energy, EA, at temperature T. Fig. 3.3 Annotate Fig. 3.3 to show the effect of adding nickel to Y and H2(g) at temperature T.

[1]
d(ii)

Area B (not labelled on Fig. 3.3) represents the number of molecules with energy greater than or equal to the activation energy when nickel is added at temperature T. State the difference, if any, between areas A and B. Explain the significance of your answer on the rate of hydrogenation of Y. Give your answer in terms of collisions.

[2]
e

Alkene Z contains two C=C bonds. Z reacts with an excess of hot concentrated acidified KMnO4 to produce only CH3COCH3, HOOCCH2COOH, CO2 and H2O. Suggest the structure of Z.

[2]
4
a(i)

Propanoic acid, CH3CH2COOH, reacts with reducing agent Q to produce propan-1-ol, CH3CH2CH2OH. Suggest the formula of reducing agent Q.

[1]
a(ii)

Complete the equation to show the reduction of propanoic acid to propan-1-ol. Use [H] to represent one atom of hydrogen from Q. ......CH3CH2COOH + ......[H] → ...................................................................................

[1]
b(i)

Propan-1-ol is converted to compound T in a three-step synthesis. In step 1, SOCl2 is added to propan-1-ol to produce compound E. In step 2, E reacts with a suitable reagent to produce butanenitrile. In step 3, butanenitrile is heated with NaOH(aq). Name the type of reaction that occurs in step 1.

[1]
b(ii)

Identify the reagent and conditions required in step 2.

[1]
b(iii)

Construct an equation to describe the reaction of butanenitrile with NaOH(aq) in step 3.

[2]
b(iv)

Name the type of reaction that occurs in step 3.

[1]
c(i)

Compounds A, B and C belong to the alcohol homologous series. Each molecule of A, B and C contains four saturated carbon atoms. Identify the type of hybridisation shown in the saturated carbon atoms of all alcohols.

[1]
c(ii)

Identify the gas produced when Na(s) is added to separate samples of each alcohol.

[1]
c(iii)

Describe the role of Na(s) when it reacts with alcohols.

[1]
d(i)

Table 4.1 shows the results of two tests on separate samples of A, B and C. Table 4.1

compoundheat under reflux with acidified K2Cr2O7warm withalkaline I2(aq)
Aremains orangeno visible change
Borange to greenno visible change
Corange to greenpale yellow precipitate

Deduce the structure of A.

[1]
d(ii)

Deduce a possible name for B.

[1]
d(iii)

Deduce the identity of C.

[1]
5

The structure of vitamin C is shown in Fig. 5.1. Fig. 5.1

Figure 1 of 6 for question 5
Figure 2 of 6 for question 5
Figure 3 of 6 for question 5
Figure 4 of 6 for question 5
Figure 5 of 6 for question 5
Figure 6 of 6 for question 5
a

Deduce the empirical formula of vitamin C.

[1]
b(i)

The mass of vitamin C present in 150.0 g of lemon is found in an experiment. stage 1 All the vitamin C in 150.0 g of lemon is extracted and dissolved in water to make 100.0 cm3 of solution L. stage 2 A 25.0 cm3 sample of solution L is titrated with 5.00×10−3 mol dm−3 I2(aq). Fig. 5.2 Exactly 36.65 cm3 of I2(aq) reacts with the 25.0 cm3 sample of solution L. [Mr: vitamin C, 176] Calculate the amount, in mol, of I2(aq) added in the titration. amount of I2(aq) = .............................. mol

[1]
b(ii)

Use your answer to (b)(i) to calculate the percentage by mass of vitamin C present in 150.0 g of lemon. (If you were unable to calculate a value for the amount of I2(aq) in (b)(i), use the value 7.65×10−4 mol. This is not the correct value.) percentage by mass = ..............................%

[2]
b(iii)

The progress of the reaction of vitamin C with I2(aq) to produce M is monitored using infrared spectroscopy. Table 5.1 indicates the presence of some absorptions in the infrared spectrum of vitamin C. Complete Table 5.1 to predict which of these absorptions, if any, are present in the infrared spectrum of M. Table 5.1

absorption/cm−1present in spectrumof vitamin Cpresent in spectrumof M
1500−1680✓
2850−2950✓
3200−3650✓

Table 5.2

bondfunctional groups containing the bondcharacteristic infrared absorption range (in wavenumbers)/cm−1
C−Ohydroxy, ester1040−1300
C=Caromatic compound, alkene1500−1680
C=Oamide1640−1690
carbonyl, carboxyl1670−1740
ester1710−1750
C≡Nnitrile2200−2250
C−Halkane2850−2950
N−Hamine, amide3300−3500
O−Hcarboxyl2500−3000
hydroxy3200−3650
[1]

Mark scheme

9701/22 · May/June 2025

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1a(i) · 1 mark

  • 1 mark

    (both) giant molecular

1a(ii) · 1 mark

  • 1 mark

    delocalised electrons move through the structure OR delocalised electrons move through / between the layers

1a(iii) · 1 mark

  • 1 mark

    no delocalised electrons (to carry charge through the structure) OR all (4) outer electrons are used in bonding

1b(i) · 2 marks

  • 1 mark

    PCl5 +4H2O → H3PO4 + 5HCl

  • 1 mark

    SiCl4 + 2H2O → SiO2 + 4HCl

1b(ii) · 2 marks

  • M1

    phosphorus(V) chloride white / yellow solid

  • M2

    silicon(IV) chloride colourless liquid

1b(iii) · 2 marks

  • M1

    (phosphorus(V) chloride) produces colourless solution

  • M2

    silicon(IV) chloride produces white suspension / white precipitate (in a colorless liquid / mixture / solution)

1c(i) · 1 mark

  • 1 mark

    species behaves as acid and base OR species reacts with (both) acid and base

1c(ii) · 1 mark

  • 1 mark

    Al2O3

1d(i) · 3 marks

  • 3 marks
    SO2–73pd
    H2O0H
    SiO21600C
    Na2O1132I
    MgO2852I

1d(ii) · 2 marks

  • M1

    not enough information

  • M2

    SO2 and / or H2O break intermolecular forces OR SO2 and / or H2O does not break covalent bonds OR only SiO2 breaks covalent bonds

2a(i) · 2 marks

  • M1

    average (weighted) mass of (an) atom(s) (of an element) OR average (weighted) mass of the isotope(s) (of an element) OR average mass of one mol of isotopes / atoms (of an element)

  • M2

    compared to the mass of the unified atomic mass unit OR on a scale in which a carbon-12 atom / isotope has a mass of exactly 12 (units) OR on a scale in which the mass of a carbon-12 atom/isotope is exactly 12 (units) OR compared to 1/12 mass of one atom of C-12 OR relative / compared to 1/12 of the mass of one mol of C-12 atoms/isotopes OR when the mass of one mol C-12 atoms / isotopes is exactly 12 (000) g

2a(ii) · 2 marks

  • M1

    relative abundance of unknown isotope = (100 – 91.9 – 6.0) = 2.1%

  • M2

    55.8 = (53.9  6.0 + 55.9  91.9 + ?  M1) / 100 = 56.9

2b · 1 mark

  • 1 mark

    1 (pair)

2c · 1 mark

  • 1 mark

    spherical / circle

2d · 2 marks

  • M1

    electrons = 23

  • M2

    neutrons = 30

2e · 2 marks

  • M1

    smaller / decreases

  • M2

    loss of outer / valence shell (of electrons) OR loss of electrons so increase in (force of) attraction from nucleus OR loss of electrons so increase nuclear attraction

3a(i) · 1 mark

  • 1 mark

    two (or more) reactants / molecules react to make (only) one product

3a(ii) · 4 marks

  • M1

    curly arrow from = of C=C to H (of HBr)

  • M2

    curly arrow from bond of Hδ+—Brδ– to Brδ– (dipole is essential)

  • M3

    correct intermediate

  • M4

    curly arrow from lone pair of Br– to C+ of intermediate drawn

3a(iii) · 2 marks

  • M1

    (2-bromopropane is made from the) more stable intermediate / carbocation

  • M2

    (due to positive) inductive effect of more alkyl groups (attached to the C+ / central carbon atom) OR (due to) greater (positive) inductive effect of 2 methyl groups (rather than 1 ethyl group)

3b(i) · 1 mark

  • 1 mark

    H2 / hydrogen AND some X / alkene left(over) / remains (at the end of this reaction)

3b(ii) · 1 mark

  • 1 mark

    link the significance of gradient values to rate AND describe either the decrease in rate or the decrease in gradient as the gradient decreases the rate decreases OR rate decreases until the reaction stops when gradient = 0

3c · 2 marks

  • 1 mark

    energy involved when all bonds are broken = (7242) AND energy involved when all bonds are made = (7490) / energy involved when all bonds are broken = (2092) AND energy involved when all bonds are made = (2340)

  • 1 mark

    energy to break bonds – energy to make bonds = –248

3d(i) · 1 mark

  • 1 mark

    EA shown on Fig. 3.3 to the left of original (on the x axis)

3d(ii) · 2 marks

  • M1

    it / (area) B larger AND rate (reaction / hydrogenation) increases OR (area) B identified by shading or labelling on Fig 3.3 AND rate (reaction / hydrogenation) increases in prose

  • M2

    increase in the frequency of effective / successful collisions OR increase in rate of effective / successful collisions

3e · 2 marks

  • M1

    H2C=C…. on one end of their molecule OR …C=C(CH3)2 on one end of their molecule OR …C=CHCH2CH=C… in the middle of their molecule

  • M2

    correct structure

4a(i) · 1 mark

  • 1 mark

    LiAlH4

4a(ii) · 1 mark

  • 1 mark

    CH3CH2CO2H + 4[H] → CH3CH2CH2OH + H2O

4b(i) · 1 mark

  • 1 mark

    substitution

4b(ii) · 1 mark

  • 1 mark

    KCN / potassium cyanide in ethanol / C2H5OH AND heat

4b(iii) · 2 marks

  • M1

    correct organic product (as molecule or anion)

  • M2

    rest of equation is correct

4b(iv) · 1 mark

  • 1 mark

    (alkaline) hydrolysis

4c(i) · 1 mark

  • 1 mark

    sp3

4c(ii) · 1 mark

  • 1 mark

    hydrogen / H2

4c(iii) · 1 mark

  • 1 mark

    (It / Na) reduces A / B / C / the alcohol (group) OR (it / Na behaves as) reducing agent OR (it / Na) donates electrons to A / B / C / the alcohol (group)

4d(i) · 1 mark

  • 1 mark

    (CH3)3COH

4d(ii) · 1 mark

  • 1 mark

    butan-1-ol OR (2) methylpropan-1-ol

4d(iii) · 1 mark

  • 1 mark

    butan-2-ol OR CH3CH2CH(OH)CH3.

5a · 1 mark

  • 1 mark

    C3H4O3

5b(i) · 1 mark

  • 1 mark

    5.00  10–3  36.65 / 1000 = 1.833  10–4 / 1.8325  10–4

5b(ii) · 2 marks

  • M1

    expression and / or correct calculation to show how the amount of vitamin C in 150 g of lemon peel is calculated 1.8325  10-4 mol  176  4 OR 7.65  10-4  176  4 = 0.5386 / 0.539

  • M2

    correct calculation based on (M1 / 150)  100 = 0.086(%) Using 1.8325  10-4 for M1, M2 = 0.086 OR 0.087(%) OR using 7.65  10-4 for M1, M2 correct calculation based on (M1 / 150)  100 = 0.359 / 0.36 (%)

5b(iii) · 1 mark

  • 1 mark
    absorption / cm–1present in spectrum of vitamin Cpresent in spectrum of M
    1500–1680✓✗ / – / blank
    2850–2950✓✓
    3200–3650✓✓

Questions

  1. 1
    Question 116 marks
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  2. 2
    Question 210 marks
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  3. 3
    Question 316 marks
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  4. 4
    Question 413 marks
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  5. 5
    Question 55 marks
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