Chemistry 9701/22 May/June 2023: question paper and mark scheme
Cambridge AS & A Level Chemistry (9701), Paper 2 AS Level Structured Questions, May/June 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
The melting points of some solids are shown in Table 1.1.
Table 1.1
solid melting point / K magnesium 923 phosphorus 317 sodium chloride 1074 sulfur 392
(a) (i) State the type of bonding present in magnesium and in sodium chloride.
bonding in magnesium ……
bonding in sodium chloride ……
[1]
(ii) Explain the difference in the melting points of magnesium and sodium chloride.
…… [1]
(iii) Explain the difference in the melting points of phosphorus and sulfur in terms of structure and bonding.
…… [2]
(b) (i) Define electronegativity.
…… [1]
(ii) Explain why electronegativity increases across a period.
…… [2]
(iii) Name the strongest intermolecular force that exists between NH3(l) molecules.
…… [1]
(iv) Draw a diagram to show the formation of the strongest intermolecular force between two molecules of NH3(l).
Include any relevant lone pairs of electrons and dipoles.
[2]
(v) The melting points of ice and ammonia are shown in Table 1.2.
Table 1.2
solid melting point / K ice 273 ammonia 195
Suggest two reasons for the difference in the melting points of ice and ammonia.
…… [2]
[Total: 12]
Mark scheme
1(a)(i) bonding in magnesium – metallic AND bonding in sodium chloride – ionic 1
1(a)(ii) bonds in NaCl are stronger than bonds in Mg 1
1(a)(iii) M1 S8 / molecules of sulfur have more electrons (than P4 / molecules of phosphorus) 2
M2 S has stronger instantaneous dipole–induced dipole forces (than phosphorus / P)
1(b)(i) power of an atom to attract electrons to itself 1
1(b)(ii) (across a period) 2
increase in nuclear charge
similar shielding
(so) increase in nuclear attraction for bonding / outer / valence electrons
OR bonding / outer / valence electron(s) are more strongly attracted to nucleus
Two correct for one mark, three correct for two marks
1(b)(iii) hydrogen bond 1
1(b)(iv) M1 link shown as a dashed line between the lone pair of electrons from N of one NH3 to one H on other NH3 2
M2 minimum 3 correct partial charges (on adjacent atoms) over two NH3 molecules
EITHER δ–N— δ+H - - - δ–N OR δ+H - - -δ–N— δ+H
1(b)(v) M1 O is more electronegative than N 2
M2 two H-bonds per water molecule : 1 per ammonia molecule.
Chlorine is a reactive element. It forms many compounds.
(a) (i) Complete Table 2.1 to show the maximum oxidation number of the elements Na to P in their chlorides.
Table 2.1
element Na Mg Al Si P
maximum oxidation number
[1]
(ii) State what determines the maximum oxidation number of elements in Period 3.
…… [1]
(b) An excess of cold water is added to the chloride of silicon.
(i) Write an equation for the reaction between an excess of cold water and the chloride of silicon.
…… [1]
(ii) Suggest the pH of the solution produced in (b)(i).
…… [1]
(c) An excess of cold water is added to the chloride of phosphorus.
(i) Write an equation for the reaction between an excess of cold water and the chloride of phosphorus.
…… [1]
(ii) Suggest the pH of the solution produced in (c)(i).
…… [1]
(d) (i) Write an equation for the reaction of chlorine with water.
…… [1]
(ii) Write an equation for the reaction of chlorine with hot NaOH(aq).
…… [1]
(e) Bleach is used as a cleaning product to kill bacteria. It is made by adding compounds like sodium chlorate(I), NaCl O, to water.
(i) Identify the formula of the ion present in bleach that kills bacteria.
…… [1]
(ii) Sodium chlorate(I), NaCl O, reacts with hydrogen peroxide to produce sodium chloride, water and oxygen gas.
Construct an equation for this reaction.
…… [1]
(iii) A sample of bleach W contains an unknown concentration of sodium chlorate(I).
10.0 cm3 of W is diluted with distilled water to make a total volume of 100 cm3 of bleach solution. 25.0 cm3 of this diluted bleach solution is added to an excess of hydrogen peroxide and the volume of gas produced measured under room conditions.
The experiment is repeated and on average 25.0 cm3 of diluted bleach solution produces
42.0 cm3 of gas.
Calculate the concentration, in g dm–3, of sodium chlorate(I) in W.
concentration of NaCl O in W = …… g dm–3 [3]
[Total: 13]
Mark scheme
2(a)(i) 1
element Na Mg Al Si P
maximum oxidation number (+)1 (+)2 (+)3 (+)4 (+)5
of elements in chlorides
All correct for one mark
2(a)(ii) number of outer/valence electrons 1
2(b)(i) SiCl4 + 2H2O → SiO2 + 4HCl 1
2(b)(ii) 1–4 1
2(c)(i) PCl5 + 4H2O → 5HCl + H3PO4 1
2(c)(ii) 1–4 1
2(d)(i) Cl2 + H2O ⇌ HOCl + HCl 1
2(d)(ii) 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O 1
2(e)(i) ClO– 1
2(e)(ii) NaClO + H2O2 → O2 + NaCl + H2O 1
2(e)(iii) M1 (show how to calculate) mol O2 gas 3
42 cm3 gas = 42 / 24000 mol oxygen gas (= 1.75 × 10–3 mol)
M2 (process to find) amount NaClO in 10 cm3 sample = 100 cm3 (diluted) bleach
M1 x 4 (= 4 × 1.75 × 10 –3 mol in 10 cm-3 = 0.007)
M3 correct calculation using M2 to find concentration
= (M2 /10) × 1000 × 74.5 = 52.2 (g / dm3)
Fig. 3.1 describes a sequence of reactions that can be used to produce a food additive, compound Y, from CH3CH2Cl .
X step 1 step 2 step 3
CH3CH2Cl CH3CH2CN CH3CH2COOH Y dilute acid Ca(s)
Fig. 3.1
(a) (i) State the reagent and conditions for step 1 in Fig. 3.1.
…… [1]
(ii) Give the systematic name of X.
…… [1]
(iii) Identify the type of reaction that occurs when dilute acid is added to X in step 2.
…… [1]
(iv) In step 3, Y and a gas are produced.
Construct an equation for step 3.
…… [2]
(b) CH3CH2COOH can also be formed from propan-1-ol and potassium dichromate(VI).
State the conditions required.
…… [1]
(c) Complete Table 3.1 to show the number of sigma bonds (σ ) and pi bonds (π) present in a molecule of X.
Table 3.1
type of bond number of bonds in X
sigma (σ )
pi (π)
[2]
[Total: 8]
Mark scheme
3(a)(i) KCN in ethanol + heat 1
3(a)(ii) propanenitrile 1
3(a)(iii) (acid) hydrolysis 1
3(a)(iv) 2 CH3CH2COOH + Ca → (CH3CH2COO)2Ca + H2 2
M1 H2 as product
M2 balanced equation based on correct formula of Y
3(b) acid (conditions) / H+(aq) AND heat under reflux 1
3(c) 2
type of bond in X number of bonds in X
sigma 8
pi 2
In industry, ethanol is made by reacting ethene with steam in the presence of H3PO4.
reaction 1 C2H4(g) + H2O(g) C2H5OH(g)
(a) Use the bond energy values in Table 4.1 to calculate the enthalpy change, ΔHr , for reaction 1.
Table 4.1
bond bond energy / kJ mol–1
C–C 350
C=C 610
C≡C 840
C–H 410
C–O 360
C=O 740
O–H 460
ΔHr = …… kJ mol–1 [2]
(b) Reaction 1 reaches equilibrium at constant temperature and pressure.
Deduce what effect increasing the pressure will have on the amount of ethanol in the new equilibrium mixture. Use Le Chatelier’s principle to explain your answer.
effect of increasing pressure ……
explanation ……
[2]
(c) The mechanism for reaction 1 can be described in three steps. Steps 1 and 2 for reaction 1 are shown in Fig. 4.1.
step 1
H 2C CH2 H H
H 2C C+ H
H
O O–
HO P OH HO P OH
O O
step 2
H H H
H3C CH2 O+ H
O
H 2C C+
H H
H
Fig. 4.1
(i) Describe the behaviour of H3PO4 in step 1 in Fig. 4.1. Explain your answer.
…… [1]
(ii) Identify the species that behaves as an electrophile in step 2 in Fig. 4.1. Explain your answer.
…… [1]
(iii) Complete Fig. 4.2 to show the mechanism for step 3 of reaction 1. Include charges, dipoles, lone pairs of electrons and curly arrows, as appropriate.
H 3C CH2 O+ H H 3C CH2 O H
H
H
O– O
HO P OH HO P OH
O O
Fig. 4.2
[2]
(iv) Describe how a catalyst affects a reaction. Explain your answer.
…… [2]
(v) Use Fig. 4.1 and Fig. 4.2 to justify why H3PO4 is described as a catalyst in reaction 1.
…… [1]
(vi) Propene also reacts with steam. A mixture of organic products is produced.
Explain why propan-2-ol is produced in the higher yield.
…… [2]
(d) Describe the covalent bonds present between the carbon atoms in an ethene molecule by completing Table 4.2.
Table 4.2
sigma (σ ) pi (π)
type of orbitals involved in bond
how the orbitals overlap
[2]
[Total: 15]
Mark scheme
4(a) Method 1 2
identification of number and type of bonds when all are broken
2O-H + 4C-H + C=C OR (2 × 460) + (4 × 410) + 610
Identification of number and type of bonds when all are made
O-H + 5C-H + C-C + C-O OR (460 + (5 × 410) + 350 + 360)
calculation of energy to break bonds – energy to make bonds
3170 – 3220 = – 50 (kJ mol–1)
Two correct for one mark, three correct for two marks.
OR
Method 2
identification of number and type of relevant bonds broken (including the new C-H only)
2O-H + C=C OR (2 × 460) + 610
identification of number and type of relevant bonds made (including the new C-H only)
O-H + C-H + C-C + C-O OR (460 + 410 + 350 + 360)
calculation using energy to break bonds – energy to make bonds
1530 – 1580 = –50 (kJ mol–1)
Two correct for one mark, three correct for two marks.
4(b) M1 amount ethanol increases 2
M2 equilibrium moves to right AND to produce fewer moles gas OR smaller amount of gas (to reduce the pressure)
4(c)(i) acid AND it donates a proton / H+ 1
4(c)(ii) (CH3CH2)+ OR CH3C+H2 AND accepts a pair of electrons 1
4(c)(iii) 2
M1 arrow from bond of one of the O-H to O of the cation
M2 arrow from lone pair on O of the anion to between H of H-O
4(c)(iv) M1 a catalyst increases rate of reaction 2
M2 by producing an alternative reaction pathway with lower activation energy
4(c)(v) it is regenerated OR it is reformed 1
4(c)(vi) M1 (propan-2-ol is produced via a) more stable carbocation OR more stable intermediate OR (CH3)2C+H is more stable 2
M2 (positive) inductive effect of more alkyl/methyl groups
OR more electron donating alkyl / methyl groups
OR greater electron donating effect of two alkyl / two methyl groups
4(d) 2
sigma pi
type of orbitals involved in bond sp2(-sp2) (2)p(-(2)p)
how the orbitals overlap direct sideways
Two correct for one mark, four correct for two marks.
(a) Describe structural isomerism.
…… [1]
(b) A and B are structural isomers with molecular formula C5H10O.
They are both straight-chained molecules with only one functional group.
Table 5.1 describes observations when separate samples of A and B are added to different reagents.
Table 5.1
reagent A B
2,4-dinitrophenylhydrazine orange precipitate appears orange precipitate appears
(2,4-DNPH reagent)
Tollens’ reagent silver mirror appears no reaction
alkaline I2(aq) no reaction no reaction
(i) Name the functional group present in both A and B.
…… [1]
(ii) Draw the structures of A and B in the boxes.
A B
[2]
(c) C is a structural isomer of A and B.
C is straight chained and has two functional groups.
C shows only one type of stereoisomerism.
Table 5.2 describes observations when separate samples of C are added to different reagents.
Table 5.2
reagent C
2,4-dinitrophenylhydrazine no reaction
(2,4-DNPH reagent)
Br2(aq) orange to colourless
alkaline I2(aq) yellow precipitate appears
(i) Draw the structure of C in the box.
C
[2]
(ii) Name the type of stereoisomerism shown by molecules of C.
…… [1]
(d) D reacts in the presence of a sulfuric acid catalyst to form E and water.
The structure of E is shown in Fig. 5.1.
E
O
O
Fig. 5.1
(i) Name the functional group present in E.
…… [1]
(ii) Identify the type of reaction that occurs when D reacts to form E.
…… [1]
(iii) Draw the structure of D in the box.
D
[1]
(iv) The infrared spectrum of E is shown in Fig. 5.2.
100
transmittance / % 50
0
4000 3000 2000 1500 1000 500 wavenumber / cm–1
Fig. 5.2
Table 5.3
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 3200–3600
Use Fig. 5.2 and Table 5.3 to predict two differences in the absorptions above 1500 cm–1 of the infrared spectrum of D compared to E. Explain your answer.
…… [2]
[Total: 12]
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
138.9 140.1 140.9 144.4 – 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
5(a) molecules / isomers with the same molecular formula / same number of atoms of each element 1
AND different structural formulae / different structures
5(b)(i) carbonyl (group) 1
5(b)(ii) 2
A B
5(c)(i) 2
C
M1 CH3CH(OH) present
M2 five carbons in a straight chain AND molecular formula C5H10O AND 1 C=C structure which shows only one type of
stereoisomerism
5(c)(ii) optical 1
5(d)(i) ester 1
5(d)(ii) condensation 1
5(d)(iii) 1
5(d)(iv) Any two of the following differences in spectra D compared to E: 2
value or range of values of
bond functional group
absorbance inclusive (cm–1)
E C=O / C double bond to O (ester) 1710–1750
D C=O (carboxy(l)) 1670–1740
D O-H OR OH bond (carboxy(l)) 2500–3000
D O-H OR OH bond (hydroxy(l) OR alcohol) 3200–3600
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