Chemistry 9701/21 May/June 2024: question paper and mark scheme
Cambridge AS & A Level Chemistry (9701), Paper 2 AS Level Structured Questions, May/June 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) The elements of Group 17 are called halogens.
Complete Table 1.1.
Table 1.1
halogen colour at 293 K
chlorine
bromine
iodine
[1]
(b) State the trend in volatility of the halogens chlorine, bromine and iodine. Explain your answer.
…… [3]
(c) Iodine is made by reacting bromine with sodium iodide.
(i) Construct an ionic equation for the reaction of bromine with sodium iodide.
…… [1]
(ii) State the role of bromine in the reaction. Explain your answer.
…… [1]
(d) Concentrated sulfuric acid is added to separate samples containing equal amounts of NaCl ,
NaBr and NaI. All three samples initially react to produce the hydrogen halide.
(i) Write an equation to describe the acid–base reaction that occurs when concentrated sulfuric acid reacts with NaBr.
…… [1]
(ii) Deduce which sodium halide, NaCl , NaBr or NaI, produces the largest percentage yield of hydrogen halide when concentrated sulfuric acid is added. Explain your answer by considering the relative reactivity of the halide ions as reducing agents.
identity of sodium halide ……
explanation ……
[3]
[Total: 10]
Mark scheme
1(a) 1
halogen colour at 293 K
chlorine green
bromine brown
iodine (dark) grey
All three correct for one mark
1(b) M1 (it / volatility) decreases 3
M2 more electrons in molecules
M3 (so) increased strength of instantaneous dipole – induced dipole forces (of attraction)
1(c)(i) Br2 + 2I– → 2Br – + I2 1
1(c)(ii) (Br2 / bromine is an) oxidising agent 1
AND
removes electron(s) from iodide (ions) I– OR increases the oxidation number of I– / iodide (ion)
1(d)(i) NaBr + H2SO4 → HBr + NaHSO4 1
OR
2NaBr + H2SO4 → 2HBr + Na2SO4
1(d)(ii) M1 sodium chloride 3
M2 Cl – is not a strong enough reducing agent to reduce H2SO4
M3 Br– & I– are strong enough reducing agents to react (further) with (the S in) H2SO4
Alternative M2 and M3
M2 ability to behave as reducing agents increases down the group
M3 EITHER Cl – is not strong enough to reduce H2SO4 OR Br – and I– react (further) with (the S in) H2SO4
(a) Sulfur chloride, SCl 2, is a liquid at room temperature. When SCl 2 is added to water, misty fumes are seen and a solution is made that turns universal indicator red.
(i) Identify the type of reaction that occurs when SCl 2 is added to water.
…… [1]
(ii) Name a chloride of a different Period 3 element that is also a liquid at room temperature and produces misty fumes when added to water.
…… [1]
(b) A molecule of SCl 2 contains two S–Cl covalent bonds.
(i) Complete the dot-and-cross diagram in Fig. 2.1 to show the arrangement of the outer electrons in a molecule of SCl 2.
Use × to show electrons from the chlorine atoms.
Use ● to show electrons from the sulfur atom.
Cl S Cl
Fig. 2.1
[2]
(ii) Predict the shape of, and bond angle in, a molecule of SCl 2 by using VSEPR theory.
shape ……
bond angle ……
[2]
(c) Solid magnesium nitride, Mg3N2, is a crystalline solid.
(i) Deduce the oxidation numbers of magnesium and nitrogen in magnesium nitride to complete Table 2.1.
Table 2.1
oxidation number in Mg3N2 magnesium nitrogen
[1]
(ii) Magnesium nitride reacts with an excess of water to produce ammonia and magnesium hydroxide only. Construct an equation to describe this reaction.
…… [1]
(iii) Explain why the solution produced in the reaction in (c)(ii) has a pH greater than 7. Refer to the products of the reaction in your answer.
…… [2]
(d) Boron nitride is a white solid that melts above 2900 °C.
Fig. 2.2 shows part of the lattice structure of a crystal of boron nitride.
boron atom nitrogen atom
van der Waals’ forces
covalent bond
Fig. 2.2
(i) Use Fig. 2.2 to deduce the empirical formula of boron nitride.
…… [1]
(ii) Suggest the identity of another crystalline solid that has atoms arranged in layers similar to that of solid boron nitride.
…… [1]
[Total: 12]
Mark scheme
2(a)(i) hydrolysis 1
2(a)(ii) silicon(IV) chloride / silicon tetrachloride 1
ALLOW phosphorus(III) chloride / phosphorus trichloride
2(b)(i) M1 1 shared pair of electrons between each S-Cl shown as 2
M2 6 non-bonding electrons around each Cl shown as AND 4 non-bonding electrons around S shown as
2(b)(ii) M1 shape non-linear 2
M2 bond angle 103–105(°)
2(c)(i) oxidation number in Mg3N2 1
magnesium (+)II / 2(+)
nitrogen –III/ 3 –
Both correct for one mark
2(c)(ii) Mg3N2 + 6H2O 3Mg(OH)2 + 2NH3 1
2(c)(iii) M1 due to OH–(aq) 2
M2 EITHER acid-base reaction of water by ammonia OR dissociation of (some) Mg(OH)2
2(d)(i) BN 1
2(d)(ii) graphite 1
(a) Define Le Chatelier’s principle.
…… [2]
(b) Reaction 1 describes the reversible reaction between yellow Fe3+(aq) and colourless
SCN–(aq) to produce red FeSCN2+(aq).
reaction 1 Fe3+(aq) + SCN–(aq) FeSCN2+(aq) ∆H = –x kJ mol–1 yellow colourless red
A mixture of Fe3+(aq), SCN–(aq) and FeSCN2+(aq) is at equilibrium at 20 °C.
The temperature of this mixture is then increased to 50 °C and allowed to reach equilibrium.
Deduce the changes that occur, if any, in the equilibrium mixture at 50 °C compared to the equilibrium mixture at 20 °C.
• change in appearance
• change in relative concentration of FeSCN2+(aq)
• change in value of the equilibrium constant, Kc
[3]
(c) In another experiment, equimolar amounts of Fe3+(aq) and SCN–(aq) are mixed together and allowed to reach equilibrium. The total volume of the mixture is 25.0 cm3.
reaction 1 Fe3+(aq) + SCN–(aq) FeSCN2+(aq)
At equilibrium the mixture contains:
• [SCN–] = 1.30 × 10–3 mol dm–3
• [FeSCN2+] = 0.300 × 10–3 mol dm–3.
(i) Calculate the initial amount, in mol, of Fe3+(aq) added to SCN–(aq) to produce this mixture.
initial amount of Fe3+(aq) = …… mol [2]
(ii) Calculate Kc for reaction 1 and state its units.
Show your working.
Kc = ……
units ……
[2]
[Total: 9]
Mark scheme
3(a) M1 (if a change in conditions occurs) the equilibrium shifts 2
M2 to minimise the change in conditions
3(b) M1 change in appearance 3
paler red / more yellow / less orange
M2 change in relative concentration of FeSCN2+(aq)
lower
M3 change in value of the equilibrium constant, Kc
lower
3(c)(i) M1 deduce the concentration of SCN– present in initial mixture 2
[SCN– eq’m] = [Fe3+ eq’m] = [original SCN-] – [FeSCN2+ produced]
1.30 × 10-3 + 0.300 × 10-3 = 1.60 × 10-3 mol dm-3
M2 find initial amount of Fe3+ initially
M1 × 25 / 1000 = amount of Fe3+ added initially = 4.00 × 10-5 (mol)
3(c)(ii) M1 value for Kc = 178 2
M2 units = mol-1 dm3
(a) Define enthalpy change of formation.
…… [2]
(b) Iron is made when iron(III) oxide is heated with carbon monoxide, as shown by reaction 2.
reaction 2 Fe2O3 + 3CO 2Fe + 3CO2
Table 4.1 shows enthalpy change of formation data measured at 298 K and 101 kPa.
Table 4.1
substance equation value for ∆H of / kJ mol–1
Fe2O3 –824.2
CO –110.5
CO2 C(s) + O2(g) CO2(g) –393.5
(i) Complete Table 4.1 by adding equations with relevant state symbols to represent:
• standard enthalpy change of formation for Fe2O3
• standard enthalpy change of formation for CO.
[2]
(ii) Use the data in Table 4.1 to calculate the enthalpy change of reaction, ∆Hr , in kJ mol–1, for reaction 2.
Show your working.
∆Hr = …… kJ mol–1 [2]
[Total: 6]
Mark scheme
4(a) M1 (enthalpy change / energy change) when one mole of a compound / substance is formed 2
M2 from its elements in their standard states
4(b)(i) equation for reaction ∆Hformation 2
Fe2O3 2Fe (s) + O2 (g) → Fe2O3 (s) –824.2
CO C (s) + O2 (g) → CO (g) –110.5
CO2 C (s) + O2 (g) CO2 (g) –393.5
4(b)(ii) ∆Hreaction = –∆HFe2O3 – (3 × –∆HCO) + (3 × –∆HCO2) 2
= –(–824.2) – (3 × –110.5) + (3 × –393.5)
= –24.8 kJ mol-1
evidence of use of correct expression and / or any rearrangement of only these bond energy values
use correct stoichiometry for these bond energy values only
calculate their expression correctly provided the expression uses only addition / subtraction and all 3 values
Hydrocarbon molecules contain covalent bonds.
(a) Define covalent bond.
…… [1]
(b) A C=C bond in an alkene is made from a σ bond and a π bond.
(i) Identify the hybridisation of the carbon atoms in a C=C bond in an alkene.
…… [1]
(ii) Draw labelled diagrams to show, in terms of orbital overlap, how the σ and π bonds are made in a C=C bond.
σ bond
π bond
[2]
(c) In electrophilic reactions involving alkenes the π bond of C=C is broken.
(i) Suggest one difference between σ and π bonds that explains why the π bond is broken in electrophilic addition reactions involving alkenes.
…… [1]
(ii) Complete Fig. 5.1 to show the mechanism for the electrophilic addition of hydrogen bromide to 2-methylpropene to produce the major organic product.
Include charges, dipoles, lone pairs of electrons and curly arrows, as appropriate.
H CH3
H C C CH3
H Br
Fig. 5.1
[4]
[Total: 9]
Mark scheme
5(a) (electrostatic) attraction between nuclei of two atoms and shared pair of electrons 1
5(b)(i) sp2 1
5(b)(ii) M1 𝜎 bond – diagram showing how 𝜎 bond is made from direct overlap of orbitals i.e. 2
OR
M2 𝜋 bond - diagram showing how 𝜋 bond is made from sideways overlap of 2× p– orbitals i.e.
OR OR
5(c)(i) EITHER (pair of) electrons in 𝜋 bond are further away from the nuclei so weaker attraction 1
OR (pair of) electrons in 𝜎 bond are closer to the two nuclei so stronger attraction
5(c)(ii) CH3 H CH3 H 4
CH3 H
H 3C C C H +
H 3C C C H
H 3C C C H
H Br H
H Br –
Br
M1 arrow from = of C=C to H of H-Br
M2 correct dipole on H-Br AND arrow from H-Br bond to Br
M3 correct intermediate AND 2-bromopropane
M4 arrow from lone pair on Br — to C+ of their intermediate (with +ve or δ+ charge)
(a) V shows stereoisomerism.
V
O
O
O
Fig. 6.1
(i) Explain what is meant by stereoisomerism.
…… [1]
(ii) Deduce the number of stereoisomers of V. Explain your reasoning.
…… [2]
(iii) Deduce the molecular formula of V.
…… [1]
(iv) Name all the functional groups present in V.
…… [1]
(b) Fig. 6.2 shows two reactions involving V.
V
O
O
O reaction 1 reaction 2
+ reagent T + reagent U
W X
OH OH
O O
O O
Fig. 6.2
(i) Identify the role of reagent T for each functional group that reacts in reaction 1.
…… [1]
(ii) Suggest the identity of reagent U in reaction 2.
…… [1]
(c) Both functional groups in one molecule of Y react with an inorganic reagent to form one molecule of Q and one molecule of methanol, CH3OH, as shown in Fig. 6.3.
Y
O
O inorganic reagent
Q + CH3OH
O
Fig. 6.3
(i) Part of the mass spectrum for Q is shown in Fig. 6.4. Only peaks with m / e greater than
198 are shown.
relative 100 abundance
x
198 199 200 201 m/e
Fig. 6.4
Calculate the relative abundance, x, of the peak at m / e = 201.
Show your working.
x = …… [2]
(ii) Q contains only hydroxyl functional groups.
Complete Table 6.1 to show the observations that occur when 2,4-dinitrophenylhydrazine
(2,4-DNPH reagent) is added to separate samples of Y and Q.
Table 6.1
observation on addition of 2,4-DNPH reagent
Y
Q
[1]
(iii) Under certain conditions, 0.0020 mol of Q reacts with an excess of sodium to produce a total of 44.8 cm3 of gas at s.t.p.
Calculate the number of hydroxyl groups present in a molecule of Q.
Show your working.
number of hydroxyl groups = …… [2]
(iv) Use Table 6.2 to describe and explain two differences between the infrared spectrum of
Y and Q in the region above 1500 cm–1.
…… [2]
Table 6.2
bond functional groups containing the bond characteristic infrared absorption range
(in wavenumbers) / cm–1
C–O hydroxy, ester 1040–1300
C=C aromatic compound, alkene 1500–1680
C=O amide 1640–1690 carbonyl, carboxyl 1670–1740 ester 1710–1750
C≡N nitrile 2200–2250
C–H alkane 2850–2950
N–H amine, amide 3300–3500
O–H carboxyl 2500–3000 hydroxy 3200–3650
[Total: 14]
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 = 6.022 × 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
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
6(a)(i) (molecules with the) same structural formula (and same molecular formula) with different arrangement of atoms / groups in 1
space
6(a)(ii) M1 number of stereoisomers = 8 2
M2 reasoning: 2 chiral centres AND 1 C=C (producing a cis-and trans or geometrical pair)
6(a)(iii) C13H20O3 1
6(a)(iv) ester AND carbonyl / ketone AND C=C (bond) / alkene 1
6(b)(i) reducing agent for C=O AND C=C 1
6(b)(ii) sodium borohydride / NaBH4 1
6(c)(i) M1 deduce n = 12 (from Y – 1C (in methanol)) 2
M2 (100 × x) ÷ (1.1 × 100) = 12 (so) x = 13.2
6(c)(ii) 1
observation on addition of 2,4–DNPH
Y orange precipitate
Q no precipitate
Both correct for one mark
6(c)(iii) ROH + Na → RONa + 1 / 2H2 2
(0.002 mol Q produced 0.002 mol H2 gas so) 2 OH groups
M1 answer indicates that OH group(s) in Q react with Na to produce the H2 in the ratio 1 mol OH : ½ mol H2
M2 uses data to show 2OH groups
6(c)(iv) M1 Y will have absorption / peak / trough between 1670–1740 due to C=O (Q will not) 2
M2 Q will have absorption / peak / trough between 3200–3600 due to O-H (Y will not)
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Chemistry 9701/21 for May/June 2024 is out of 60 marks, with 1 hour 15 minutes to answer it, across 6 questions.
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