Chemistry 9701/22 October/November 2025: question paper and mark scheme
Cambridge AS & A Level Chemistry (9701), Paper 2 AS Structured Questions, October/November 2025. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
Manganese, Mn, and its compounds are widely used in many chemical reactions.
(a) Mn is usually found as a single isotope, manganese-55.
(i) Determine the number of protons, neutrons and electrons in an atom of manganese-55.
number of protons …… neutrons …… electrons ……
[1]
(ii) Define isotopes.
…… [1]
(iii) A sample of manganese from the Moon is found to contain manganese-53 in addition to manganese-55.
State the two pieces of information needed to determine the relative atomic mass, Ar , of manganese in this sample.
1 ……
2 ……
[2]
(b) The shorthand electronic configuration of manganese is [Ar] 3d5 4s2.
(i) Complete the full electronic configuration of manganese.
…… 3d5 4s2 [1]
(ii) Deduce the total number of unpaired electrons in an atom of manganese.
…… [1]
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(c) Manganese(IV) oxide reacts with methanal, CH2O, in acidic conditions to produce carbon dioxide. The movement of electrons to or from relevant species is shown in the following half-equations.
half-equation 1 CH2O + H2O CO2 + 4H+ + 4e–
half-equation 2 MnO2 + 4H+ + 2e– Mn2+ + 2H2O
(i) Identify the species that is reduced in half-equation 2. Explain your answer.
…… [1]
(ii) The oxidation state of carbon in methanal is 0.
Calculate the oxidation state of carbon in carbon dioxide.
…… [1]
(iii) Construct the ionic equation for the reaction of manganese(IV) oxide with methanal in
acidic conditions.
…… [2]
[Total: 10]
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4
Mark scheme
1(a)(i) 25 protons 30 neutrons 25 electrons 1
1(a)(ii) (atoms of an element) with the same number of protons and different number of neutrons 1
1(a)(iii) M1 (relative) isotopic mass. 2
OR (relative isotopic) mass of Mn-55 and Mn-53
M2 (relative) abundance of each isotope
1(b)(i) 1s2 2s2 2p6 3s2 3p6 (3d5 4s2) 1
1(b)(ii) 5 1
1(c)(i) Mn / MnO2 AND it gains electrons 1
1(c)(ii) (+)4 1
1(c)(iii) Due to an issue with question 1(c)(iii), the question has been removed from the question paper. N/A
The Period 3 elements show trends in physical and chemical properties across the period.
(a) (i) Explain why the elements Na to Al are good electrical conductors.
…… [1]
(ii) Explain why the elements P, S and Cl do not conduct electricity.
…… [1]
(b) Fig. 2.1 shows the variation in melting point of the Period 3 elements Si to Cl.
1800
1600
1400
1200
melting 1000
point / K 800
600
400
200
0
Si P S Cl
Fig. 2.1
The Period 3 elements Si to Cl are all non-metals.
Explain why there is a large difference between the melting point of Si and the melting points of P, S and Cl.
…… [2]
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(c) Table 2.1 gives some information about some Period 3 chlorides.
Row B refers to the pH of the solution that forms when the Period 3 chloride is added to water.
Table 2.1
formula of Period 3 chloride NaCl MgCl 2 Al Cl 3 SiCl4 PCl 5 oxidation number of
A element bonded to Cl
B pH of solution 6.5
C bonding ionic
D structure giant
Complete Table 2.1.
You may use the following abbreviations.
I = ionic, C = covalent, M = metallic
G = giant, S = simple
[4]
(d) (i) Write an equation for the formation of Al Cl 3 from its elements.
…… [1]
(ii) Write an equation for the formation of H3PO4 from PCl 5.
…… [1]
(e) S2Cl2(l) reacts with Cl2(g) in a reversible reaction to form SCl2(l). Under certain conditions, a dynamic equilibrium is established.
S2Cl2(l) + Cl2(g) 2SCl2(l) ΔH = – 41 kJ mol–1
(i) State what is meant by dynamic equilibrium.
…… [1]
(ii) Identify the condition necessary to establish dynamic equilibrium.
…… [1]
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(iii) S2Cl2(l) is yellow and SCl2(l) is red.
S2Cl2(l) + Cl2(g) 2SCl2(l) ΔH = – 41 kJ mol–1
State what is observed when the following changes are made to an equilibrium mixture of S2Cl2(l) and SCl2(l).
Explain your answers.
• The equilibrium mixture is warmed gently.
observation ……
explanation ……
• The overall pressure of the equilibrium mixture is increased.
observation ……
explanation ……
[4]
(f) Aqueous MgCl 2 reacts with aqueous Na2CO3 to form a white precipitate. Upon heating, the white precipitate undergoes thermal decomposition.
(i) Construct an equation to show the reaction of aqueous MgCl 2 with aqueous Na2CO3.
Use state symbols in your equation.
…… [2]
(ii) Identify the products of the thermal decomposition of the white precipitate.
…… [1]
(iii) State the trend in thermal stability of the Group 2 carbonates down the group.
…… [1]
[Total: 20]
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Mark scheme
2(a)(i) delocalised electrons move (through the structure / between layers) 1
2(a)(ii) P / S / Cl have no charge carriers 1
OR (no ions and) no delocalised electrons
OR no mobile / moving ions
2(b) M1 Si has (lots of) strong covalent bonds broken 2
M2 P / S / Cl (only) has weak Van der Waals’ (forces) broken
2(c) M1 4
+1 +2 +3 +4 +5
M2 7 6.5 0–4 0–4 0–4
M3 I I I C C
M4 G G G S S
2(d)(i) 2Al + 3Cl2 → 2AlCl3 1
2(d)(ii) PCl5 + 4H2O → H3PO4 + 5HCl 1
2(e)(i) rate of forward and backward (reaction) are the same / equal 1
OR (ratio of) concentrations of reactants and products are constant
2(e)(ii) closed system 1
2(e)(iii) M1 (mixture / solution becomes more) yellow 4
M2 (forward) reaction is exothermic (equilibrium shifts left)
OR equilibrium / reaction shifts (left) to absorb (additional) heat / energy
M3 (mixture / solution becomes more) red
M4 fewer moles of gas on right-hand side
OR (gas on left and) no gas on the right
2(f)(i) MgCl2(aq) + Na2CO3(aq) → MgCO3(s) + 2NaCl(aq) 2
M1 formulae in balanced equation
M2 state symbols
2(f)(ii) MgO / magnesium oxide AND CO2 / carbon dioxide 1
2(f)(iii) (thermal stability) increases (down the group) 1
The alkanes are a homologous series of organic molecules. Alkanes are generally unreactive and are commonly used as fuels.
(a) Define homologous series.
…… [2]
(b) Give two reasons to explain the general unreactivity of alkanes.
1 ……
2 ……
[2]
(c) Alkanes with low relative molecular mass, Mr , are more useful than those found in heavier crude oil fractions.
Name the process that is used to obtain alkanes with low Mr from heavier crude oil fractions.
…… [1]
(d) Hexane, C6H14, has four structural isomers.
Fig. 3.1 shows hexane and two of its structural isomers.
hexane A B C D
Fig. 3.1
(i) Complete Fig. 3.1 by drawing structures for C and D, the other two structural isomers of hexane. [2]
(ii) A, B and hexane have different boiling points.
Arrange A, B and hexane in order of increasing boiling point.
Explain your answer.
lowest …… < …… < …… highest
[3]
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(e) Hexane can be converted into compounds E and F at high temperature and pressure. Fig. 3.2 shows the reaction scheme involving hexane, E and F.
∆H1 = +215 kJ mol−1 ∆H2 reaction 1 reaction 2
E F
reaction 3
Fig. 3.2
(i) Identify a suitable reagent for reaction 3.
…… [1]
(ii) Use the data in Fig. 3.2 and in Table 3.1 to calculate the enthalpy change of reaction 2,
ΔH2.
Table 3.1
enthalpy change of compound formation, ∆Hf / kJ mol−1
−156
+48
ΔH2 = …… kJ mol–1 [2]
(iii) Write an equation for the complete combustion of hexane.
…… [1]
[Total: 14]
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Mark scheme
3(a) M1 (family of molecules / compounds / series with) same general formula 2
OR (family of molecules where) molecular formula differs (only) by (units of) CH 2(n).
M2 similar chemical properties / reactions
3(b) M1 strong C—H (bonds) 2
M2 non-polar (C—H / molecule / bonds)
3(c) cracking 1
3(d)(i) 2
M1
M2
3(d)(ii) M1 B < A < hexane. 3
M2 hexane / unbranched has stronger / more IMFs
M3 hexane / unbranched requires most / more energy
3(e)(i) H2 ((g)) OR hydrogen (gas) 1
3(e)(ii) M1 +215 + (–156) – (+48) = 2
M2 +11 kJ mol–1
3(e)(iii) C6H14 + 9½O2 → 6CO2 + 7H2O 1
Fig. 4.1 shows how propane, C3H8, can be converted to propanoic acid, CH3CH2COOH.
reaction 1 reaction 2 reaction 3
C3H8 CH3CH2CH2Cl CH3CH2CH2OH CH3CH2COOH
Fig. 4.1
(a) Reaction 1 in Fig. 4.1 takes place in the presence of sunlight.
C3H8 + Cl 2 CH3CH2CH2Cl + HCl
The reaction takes place via initiation, propagation and termination steps.
(i) Name the mechanism shown by reaction 1.
…… [1]
(ii) Complete the mechanism for reaction 1.
Construct equations to describe the steps of the mechanism.
initiation Cl 2 → 2Cl•
propagation 1 ……
propagation 2 ……
termination …… CH3CH2CH2Cl
[3]
(iii) Reaction 1 is initiated by the bond fission of Cl 2.
State the type of bond fission shown in the initiation step.
…… [1]
(iv) Compound Q is a by-product of reaction 1.
Q
H Cl H
H C C C Cl
H H H
Name Q.
…… [1]
* 0000800000011 * DFD
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(v) The molecular formula of Q is C3H6Cl 2.
Identify the types of structural isomerism and stereoisomerism that a molecule with molecular formula C3H6Cl 2 can show.
type of structural isomerism ……
type of stereoisomerism ……
[2]
(b) State the reagent and solvent required for reaction 2.
…… [1]
(c) Reaction 3 takes place when CH3CH2CH2OH is heated under reflux with acidified potassium dichromate(VI) solution.
(i) State the colour change that takes place in the reaction mixture.
…… [1]
(ii) Construct an equation to represent reaction 3. Use [O] to represent an atom of oxygen from the oxidising agent.
…… [1]
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(d) CH3CH2COOH reacts with an unsaturated alcohol R to form unsaturated ester S.
(i) State the type of reaction that forms S.
…… [1]
(ii) The infrared spectrum of S is shown in Fig. 4.2.
100
transmittance / % 50
0
4000 3000 2000 1500 1000 500
wavenumber / cm−1
Fig. 4.2
Three absorptions in the infrared spectrum in Fig. 4.2 confirm that S is an ester and is unsaturated.
• Write 1, 2 or 3 on Fig. 4.2 against each of these three absorptions.
• Complete Table 4.1 to show which bond is responsible for each absorption that you have identified in Fig. 4.2.
Table 4.1
absorption 1 2 3 bond responsible
[3]
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Table 4.2
characteristic infrared absorption range bond functional groups containing the bond
(in wavenumbers) / cm–1
C–O hydroxy, ester 1040–1300
C=C aromatic compound, alkene 1500–1680 amide 1640–1690
C=O carbonyl, carboxyl 1670–1740 ester 1710–1750
C N nitrile 2200–2250
C–H alkane 2850–2950
N–H amine, amide 3300–3500 carboxyl 2500–3000
O–H hydroxy 3200–3650
(iii) The mass spectrum of S shows the following peaks.
Table 4.3
peak relative abundance
M+ 4.7
[M+1]+ 0.31
Use Table 4.3 to calculate the number of carbon atoms in S.
number of carbon atoms in S = …… [1]
[Total: 16]
DFD
14
* 0000800000015 * DFD
15
Important values, constants and standards
molar gas constant R = 8.31 J K–1 mol–1
Faraday constant F = 9.65 × 104 C mol–1
Avogadro constant L = 6.02 × 1023 mol–1 electronic charge e = –1.60 × 10–19 C molar volume of gas Vm = 22.4 dm3 mol–1 at s.t.p. (101 kPa and 273 K)
Vm = 24.0 dm3 mol–1 at room conditions
ionic product of water Kw = 1.00 × 10–14 mol2 dm–6 (at 298 K (25 °C)) specific heat capacity of water c = 4.18 kJ kg–1 K–1 (4.18 J g–1 K–1)
The Periodic Table of Elements
Group
1 2 13 14 15 16 17 18
1 2
H He hydrogen helium
Key 1.0 4.0
3 4 atomic number 5 6 7 8 9 10
Li Be atomic symbol B C N O F Ne lithium beryllium name boron carbon nitrogen oxygen fluorine neon
6.9 9.0 relative atomic mass 10.8 12.0 14.0 16.0 19.0 20.2
11 12 13 14 15 16 17 18
Na Mg Al Si P S Cl Ar sodium magnesium aluminium silicon phosphorus sulfur chlorine argon
DFD
23.0 24.3 3 4 5 6 7 8 9 10 11 12 27.0 28.1 31.0 32.1 35.5 39.9
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
at www.cambridgeinternational.org after the live examination series.
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr potassium calcium scandium titanium vanadium chromium manganese iron cobalt nickel copper zinc gallium germanium arsenic selenium bromine krypton
39.1 40.1 45.0 47.9 50.9 52.0 54.9 55.8 58.9 58.7 63.5 65.4 69.7 72.6 74.9 79.0 79.9 83.8
37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
16
Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe rubidium strontium yttrium zirconium niobium molybdenum technetium ruthenium rhodium palladium silver cadmium indium tin antimony tellurium iodine xenon
85.5 87.6 88.9 91.2 92.9 95.9 – 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3
55 56 57–71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 lanthanoids
Cs Ba Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn caesium barium hafnium tantalum tungsten rhenium osmium iridium platinum gold mercury thallium lead bismuth polonium astatine radon
132.9 137.3 178.5 180.9 183.8 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 – – –
87 88 89–103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 actinoids
Fr Ra Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv Ts Og francium radium rutherfordium dubnium seaborgium bohrium hassium meitnerium darmstadtium roentgenium copernicium nihonium flerovium moscovium livermorium tennessine oganesson
– – – – – – – – – – – – – – – – –
57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 lanthanoids La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu lanthanum cerium praseodymium neodymium promethium samarium europium gadolinium terbium dysprosium holmium erbium thulium ytterbium lutetium
138.9 140.1 140.9 144.2 – 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.1 175.0
89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 actinoids Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr actinium thorium protactinium uranium neptunium plutonium americium curium berkelium californium einsteinium fermium mendelevium nobelium lawrencium
– 232.0 231.0 238.0 – – – – – – – – – – –
Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download
To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge
Mark scheme
4(a)(i) free-radical substitution 1
4(a)(ii) M1 C3H8 + Cl● → CH3CH2CH2● + HCl 3
OR C3H8 + Cl● → C3H7● + HCl
M2 CH3CH2CH2● + Cl2 → CH3CH2CH2Cl + Cl●
OR C3H7● + Cl2 → C3H7Cl + Cl● etc
M3 CH3CH2CH2● + Cl● → CH3CH2CH2Cl
OR C3H7● + Cl● → CH3CH2CH2Cl
4(a)(iii) homolytic 1
4(a)(iv) 1,2-dichloropropane 1
4(a)(v) M1 (type of structural isomerism) positional 2
M2 (type of stereoisomerism) optical
4(b) sodium hydroxide AND water 1
4(c)(i) orange to green 1
4(c)(ii) CH3CH2CH2OH + 2[O] → CH3CH2COOH + H2O 1
4(d)(i) condensation 1
4(d)(ii) M1 absorption labelled at approx. 1750 cm–1 identified as C=O 3
M2 absorption labelled at approx. 1650 cm–1 identified as C=C
M3 absorption labelled at approx. 1170 cm–1 identified as C—O
4(d)(iii) (0.31 / 4.7 × 100 / 1.1) = 6 carbon atoms 1
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Chemistry 9701/22 for October/November 2025 is out of 60 marks, with 1 hour 15 minutes to answer it, across 4 questions.
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