Chemistry 9701/23 October/November 2023: question paper and mark scheme
Cambridge AS & A Level Chemistry (9701), Paper 2 AS Structured Questions, October/November 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
The elements phosphorus, sulfur and chlorine are in Period 3 of the Periodic Table.
Table 1.1 shows some properties of the elements P to Cl .
The first ionisation energy of S is not shown.
Table 1.1
property P S Cl number of electrons in 3p subshell total number of unpaired electrons first ionisation energy / kJ mol–1 1060 1260 formula of most common anion P3– S2– Cl –
(a) (i) Complete Table 1.1 to show the number of electrons in the 3p subshell and the total number of unpaired electrons in an atom of P, S and Cl . [2]
(ii) Construct an equation to represent the first ionisation energy of P.
…… [1]
(iii) Three possible values for the first ionisation energy of S are given.
1000 kJ mol–1 1160 kJ mol–1 1320 kJ mol–1
Circle the correct value.
Explain your choice by comparing your chosen value to those of P and Cl .
…… [4]
(b) P3–, S2– and Cl – have the same number of electrons.
(i) Give the full electronic configuration of P3–.
…… [1]
(ii) State the trend in ionic radius shown by P3–, S2– and Cl –.
Explain your answer.
…… [2]
(c) A student does three tests on separate samples of NaCl (aq).
Complete Table 1.2 with the observations the student makes in each test.
Table 1.2
test test observations
1 addition of a few drops of
Br2(aq)
2 addition of a few drops of concentrated H2SO4
3 addition of a few drops of dilute AgNO3(aq)
[3]
(d) POCl3 shows similar chemical properties to PCl5.
POCl3 has a melting point of 1°C and a boiling point of 106 °C.
POCl3 reacts vigorously with water, forming misty fumes and an acidic solution.
(i) Explain how the information in (d) suggests the structure and bonding of POCl3 is simple covalent.
…… [2]
(ii) Construct an equation for the reaction of POCl3 with water.
POCl3 + …… ……
[1]
(iii) POCl3 contains a double covalent bond between P and O.
Complete the dot-and-cross diagram, in Fig. 1.1, to show the bonding in POCl3.
Show outer shell electrons only.
O
Cl P Cl
Cl
Fig. 1.1
[2]
(e) POCl3(g) forms when PCl3(g) reacts with O2(g).
2PCl3(g) + O2(g) 2POCl3(g)
Table 1.3 gives some relevant data.
Table 1.3
process value / kJ mol–1 enthalpy change of formation of PCl3(g) –289 enthalpy change of formation of POCl3(g) –592
O2(g) 2O(g) +496
(i) Define enthalpy change of formation, ΔHf .
…… [2]
(ii) Calculate the bond energy of P=O in POCl 3 using the data in Table 1.3.
Show your working.
bond energy of P=O = …… kJ mol–1
[2]
[Total: 22]
Mark scheme
1(a)(i) total # 3p e– 3 4 5 1
total # unpaired e– 3 2 1 1
1(a)(ii) P(g) → P+(g) + e(–) 1
1(a)(iii) 1000 (kJmol−1) 1
(S) less / least / lowest nuclear attraction (to outer electrons) (than P OR Cl) 1
OR
nuclear attraction for Cl is stronger (compared to S (and or P))
S less nuclear charge than Cl 1
OR
Cl has a greater nuclear charge than S
(S has a greater nuclear charge (of outer electrons) than P BUT) 1
S has two electrons in a (3)p orbital AND resulting in spin-pair repulsion
1(b)(i) 1s2 2s2 2p6 3s2 3p6 1
1(b)(ii) ● decreases (from P3− to Cl−) 2
● increased attraction (of outer e−) for nucleus owtte
● increased nuclear charge
● same shielding
●√●√
1(c) (Br2) no visible reaction 1
OR
colourless to orange / brown / yellow solution
(conc. H2SO4) no visible reaction 1
OR
Solution remains colourless
(dil. AgNO3) white precipitate forms 1
1(d)(i) (relatively low melting and boiling points suggests) weak intermolecular / VdW forces 1
(vigorous reaction with water suggests) hydrolysis 1
1(d)(ii) POCl3 + 3H2O → H3PO4 + 3HCl 1
1(d)(iii) 1
Correct bonding e−
rest of electrons correct (i.e. to make 32 electrons in total) 1
1(e)(i) enthalpy change when one mole of a compound / substance is formed 1
from its constituent elements in their standard states 1
1(e)(ii) 2(–592) = 2(–289) + (496) – 2(P=O) 1
OR
(2) P=O = 2(–289) + (496) + 2(592)
EP= O = ½(1102) = +551 (kJ mol−1) 1
Barium hydroxide, Ba(OH)2 , is a strong base used in inorganic and organic reactions.
Fig. 2.1 shows a reaction scheme involving Ba(OH)2.
H2O(l)
BaO(s) Ba(OH)2(aq) BaCO3(s) + H2O(l) reaction 1 reaction 2
CH3COOH(aq) reaction 3
A(aq) + H2O(l)
Fig. 2.1
(a) (i) State the variation in solubilities of group 2 hydroxides.
…… [1]
(ii) State what is observed in reaction 1.
…… [1]
(iii) Suggest a reactant for reaction 2.
…… [1]
(iv) Identify A.
…… [1]
(v) Ba(OH)2 is made by the reaction of Ba with water.
Write an equation for this reaction.
…… [1]
(b) The mineral barytocalcite contains both BaCO3 and CaCO3. Both compounds decompose on heating.
(i) State which compound decomposes first when barytocalcite is heated.
Explain your answer.
…… [1]
(ii) Construct an equation for the complete thermal decomposition of barytocalcite.
The formula of barytocalcite is BaCa(CO3)2.
BaCa(CO3)2 …… [1]
(c) Ba(OH)2 is used to hydrolyse organic compounds.
Fig. 2.2 shows the reaction of B with Ba(OH)2, followed by acidification.
Draw the structures of the organic products of the process shown in Fig. 2.2.
B
CO2CH3
CN
Ba(OH)2 then HNO3
organic product 1 organic product 2
+ + Ba(NO3)2
Fig. 2.2
[3]
[Total: 10]
Mark scheme
2(a)(i) increases down the group 1
2(a)(ii) white solid disappears 1
2(a)(iii) CO2 / carbon dioxide 1
2(a)(iv) Ba(CH3COO)2 / barium ethanoate 1
2(a)(v) Ba + 2H2O → Ba(OH)2 + H2 1
2(b)(i) CaCO3 AND thermal stability of carbonates increases down the group 1
2(b)(ii) BaCa(CO3)2 → BaO + CaO + 2CO2 1
2(c) CH3OH 1
1
one COOH group AND correct carbon framework (5C’s)
(each C has the correct valency)
both COOH groups 1
Potassium chlorate, KCl O3, is widely used as an oxidising agent and to make O2(g).
(a) Define oxidising agent.
…… [1]
(b) KCl O3(s) decomposes when heated.
MnO2(s) catalyses the exothermic decomposition reaction.
Complete and label the diagram in Fig. 3.1 to show the effect of MnO2(s) on the decomposition of KCl O3(s).
enthalpy
reactants
products
progress of reaction
Fig. 3.1
[2]
(c) When KCl O3 is heated without a catalyst, KCl O4 and KCl form.
4KCl O3 3KCl O4 + KCl
Explain why this reaction is described as a disproportionation reaction.
…… [1]
(d) Molten KCl O3 reacts with glucose, C6H12O6.
4KCl O3 + C6H12O6 6CO2 + 6H2O + 4KCl
KCl O3 melts at 630 K. At this temperature, both CO2 and H2O are gases.
(i) Use the ideal gas equation to calculate the volume, in m3, of one mole of gas at 630 K and 1.00 × 105 Pa.
Show your working. Give your answer to 3 significant figures.
volume of 1 mol of gas = …… m3
[1]
(ii) 5.00 g of C6H12O6 reacts completely with molten KCl O3.
Use your answer to (d)(i) to calculate the total volume of gas released at 630 K and
1.00 × 105 Pa in this reaction.
(If you were unable to answer (d)(i), use 0.0463 m3 in this question. This is not the correct answer to (d)(i).)
total volume of gas released = …… m3
[2]
(e) The structure of glucose, C6H12O6, is shown in Fig. 3.2.
glucose
OH OH
HO
O
OH OH
Fig. 3.2
(i) Complete Table 3.1 to identify the number of primary, secondary and tertiary alcohol groups present in the structure shown in Fig. 3.2.
Table 3.1
type of alcohol group primary secondary tertiary
number of groups
[1]
(ii) Separate samples of aqueous glucose are tested with the reagents shown in Table 3.2.
Complete Table 3.2 with the observation for each reaction.
Write “no reaction” if applicable.
Table 3.2
reagent and conditions observation with glucose
acidified KMnO4(aq) and warm
Fehling’s reagent and warm
alkaline I2(aq) and warm
[3]
(iii) There are many structural isomers of C6H12O6.
Define structural isomers.
…… [1]
[Total: 12]
Mark scheme
3(a) compound / molecule / substance that oxidises another / is reduced 1
3(b) 1
uncatalysed hump
catalysed hump lower than uncatalysed 1
3(c) Cl / chlorine is both oxidised and reduced 1
3(d)(i) pV = nRT ∴ V = 8.31 × 630 ÷ 100 000 = 0.0524 (m3) 1
3(d)(ii) moles of gas = 12 × 5.00 / 180 1
volume of gas = M1 × (d)(i) = 0.0175 (m3) 1
3(e)(i) 1 4 0 1
3(e)(ii) Solution turns from purple to colourless 1
Orange / red precipitate forms 1
no reaction 1
3(e)(iii) molecules with the same molecular formula but different structural formulae 1
4(a(i) 3-methylbut-1-ene 1
Compounds C and D are alkenes with the same molecular formula, C5H10.
C D
Fig. 4.1
(a) (i) Give the systematic name of D.
…… [1]
(ii) Explain why C and D do not show geometrical (cis/trans) isomerism.
…… [1]
(iii) Draw the structure of a molecule that is a positional isomer of C and D.
[1]
(iv) Give the structural formula of the compound formed when D reacts with H2(g) in the presence of a Pt catalyst.
…… [1]
(v) C can form an addition polymer.
Draw the structure of one repeat unit of this addition polymer.
[1]
(b) The mass spectrum of C shows a molecular ion peak at m/e = 70. This peak has a relative intensity of 48.7.
The relative intensity of the [M+1] peak is 2.7.
Show that this information is consistent with the molecular formula of C.
[2]
(c) C and D both react with HBr.
(i) C reacts with HBr to form E.
Complete the diagram in Fig. 4.2 to show the mechanism for this reaction.
Draw the structure of the organic intermediate.
Include charges, dipoles, lone pairs of electrons and curly arrows, as appropriate.
C E
Br
H – Br
Fig. 4.2
[3]
(ii) D reacts with HBr to produce F, a chiral bromoalkane.
Draw the structure of F.
F
[1]
(iii) Explain why the reaction of HBr with C and D produces different major products.
C D
Fig. 4.3
…… [2]
(d) C can be used to form H.
H
O
O
Fig. 4.4
One possible synthesis of H is shown in Fig. 4.5. Different portions of C are used in reactions
1 and 3. Some of the products are then combined to produce H.
Fig. 4.5 does not show any of the inorganic products of the reactions.
C E
HBr(g)
J reaction 1 reaction 2
Br
reaction 3
O
G +
H
O then: J + G reaction 4
O
Fig. 4.5
Complete Table 4.1 with the reagents and conditions required for each of the reactions shown in Fig. 4.5.
Table 4.1
reagent and conditions
reaction 1 C E HBr(g)
reaction 2 E J
O reaction 3 C G+
reaction 4 J+G H
[3]
[Total: 16]
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
20
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
138.9 140.1 140.9 144.4 – 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)(ii) one end / one C of the C=C bond has the same groups attached 1
4(a)(iii) 1
4(a)(iv) (CH3)2CHCH2CH3 1
4(a)(v) 1
4(b) 2.7 100 1
× = 5(.04)
48.7 1.1
Mr = 5 × 12 + 10 × 1 = 70 1
4(c)(i) 1
correct dipole on H-Br AND curly arrow from C=C bond to Hσ+ AND curly arrow from H—Br bond to Brσ–
correct (Markovnikoff) intermediate 1
curly arrow from lone pair on Br− to C+ of intermediate 1
4(c)(ii) 1
4(c)(iii) ●C has a tertiary carbocation and a 2ndry carbocation 2
●D has a secondary carbocation and a primary carbocation
● alkyl groups have a +ve / positive / +I / inductive effect
● more alkyl groups / more inductive effects give a more stable intermediate / carbocation / C+
● In C tertiary carbocation is more stable than the secondary carbocation
Any two ● for 1 mark
Any four ● for 2 marks
4(d) (reaction 2) NaOH(aq) 1
(reaction 3) hot acidified concentrated KMnO4 1
(reaction 4) conc H2SO4 catalyst 1
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Chemistry 9701/23 for October/November 2023 is out of 60 marks, with 1 hour 15 minutes to answer it, across 4 questions.
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