Ch 1: Energetics and Group 2 | Marking Scheme #1.

1
M1 Pt Ag Ag+
M2 gas delivery system H2 H+
M3 voltmeter salt bridge wiring Pt to V to Ag

4b(ii) E value would be more negative AND shifts Ag+ (+ e–) ⇋ Ag to the left 1
1 2024 May/Jun
4(c)
| Paper 41 | Q4de
enthalpy change when one mole of a solute dissolves in water 1

4(d)(i) 2

9701/41 M1 two arrows in blue Cambridge International AS & A Level – Mark Scheme May/June 2024
M2 correct species in red with all state symbols PUBLISHED

© Question
Cambridge University Press & Assessment 2024 Answer
Page 8 of 18 Marks

4(d)(ii) Hlatt = –811.6 (kJ mol–1) 1

4(e) M1 Mg(NO3)2(s), NaNO3(s), RbNO3(s) 3

M2 Mg2+ has a higher charge than Na+ OR Rb+

AND Na+ has a smaller radius than Rb+

M3 correct statement relating magnitude of LE to attraction between ions OR strength of ionic bonds

Question Answer Marks

9701/42 Cambridge International AS & A Level – Mark Scheme May/June 2024
PUBLISHED
5(a)(i) tangent drawn at t = 0 AND gradient of tangent calculated at t = 0 1
2 2024 May/Jun | Paper 42 | Q1
Question answer between 0.016 to 0.040 Answer Marks
5(a)(ii)
1(a) [I–] magnesium
M1 stays constant / [I–] does
> calcium not change
> strontium 14
OR H
M2 thelattoverall
and H order is one under these conditions
hyd both become less exothermic / less negative
M3 Hlatt changes less OR Hhyd is dominant factor
5(b) M1 equation 1: 2Fe2+ + S2O82– → 2Fe3+ + 2SO42– 2
M4 Hsol becomes less exothermic / less negative / more positive / more endothermic
M2 equation 2: 2Fe3+ + 2I– → 2Fe2+ + I2
1(b) M1H / energy change when 1 mole of an ionic solid / compound is formed 2
5(c) the rate constant and rate of reaction 1 will both increase 1
M2 from gaseous ions (under standard conditions)
5(d) k = 0.693 ÷ t½ 1
1(c) M1 as ionic radii increases AND Hlatt less exothermic 2
t½ = 0.693 ÷ 0.0158 = 43.9 s
M2 as ionic charge increases AND Hlatt increases/more exothermic
5(e) NO + Br2 → NOBr2 OR NO + Br2 → NOBr + Br 2
1(d)(i) NOBr2 + NO → 2NOBr OR Br + NO → NOBr 3

M1 two equations adding up to the overall equation

M2 step 1 has one NO and one Br2 only and is identified as slow step

© Cambridge University Press & Assessment 2024 Page 9 of 18

9701/42 Cambridge International AS & A Level – Mark Scheme May/June 2024

any two [1] any three [2] all four [3] PUBLISHED
Question Answer Marks
9701/42 Cambridge International AS & A Level – Mark Scheme May/June 2024
1(d)(ii) PUBLISHED
M1 selection of ONLY six correct values (–381, 89, 419, 279, –200, 640) AND use of  2 as only multiplier with K 2
Question M2 correct evaluation of data used ecf Answer Marks

© Cambridge –381
1(d)(ii) University
M1 = (89
selection
Press  of
2) ONLY
+ (419
& Assessment  2)correct
six
2024 + 279 values
+ (–200)
(–381, + H
+ 64089, o
419,latt
279,5 of–200,
Page 15 640) AND use of  2 as only multiplier with K 2
Ho
latt = –2116
M2 correct (kJ mol–1
evaluation of) data used ecf

–381 = (89  2) + (419  2) + 279 + (–200) + 640 + Holatt

Question Holatt = –2116 (kJ mol–1) Answer Marks

2(a)(i) 2LiNO3 → Li2O + 2NO2 + ½O2 1

3 2024 May/Jun | Paper 42 | Q2a
Question Answer Marks
2(a)(ii) M1 radius / size of cation / M+ increases OR charge density of ion decreases 2
2(a)(i) M2 less polarisation /
2LiNO → Li O + 2NO + ½O distortion of anion / nitrate ion / NO3 / less weakening of NO bond
–
1
3 2 2 2

2(b)
2(a)(ii) M1 M2
M1 any/ size
radius two for one mark
of cation / M+ or all four for
increases ORtwo marks:
charge density of ion decreases 3
2
 less
M2 mol polarisation 0.125  0.0500
total MnO4– /=distortion of anion= /6.25  10
nitrate
–3
ion / NO3– / less weakening of NO bond
 mol Fe = 0.0400  0.0225 = 9.00  10
2+ –4

2(b)  M2
M1 molany
unreacted MnOmark
two for one –
or all four
4 = 9.00 10–4for
÷ 5two
= 1.80  10–4 ecf
marks: 3
– = 6.25  10–3 - 1.80  10–4–3
 mol totalreacted
MnOMnO
4 =4 0.125
–  0.0500 = 6.25  10 = 6.07  10–3 ecf
 mol Fe2+ = 0.0400  0.0225 = 9.00  10–4
M3 mol unreacted MnO4– = 9.00  10–4 ÷ 5 = 1.80  10–4 ecf
– = 2.5  6.07 –
mol NO MnO4 =106.25  10–3 - 1.80
–3 = 1.5175 10–2 10–4 = 6.07  10–3 ecf
mol2 reacted
conc NaNO2 = 4  1.5175  10–2 = 6.07-6.08  10–2 mol dm–3 ecf min 2sf
M3
2(c)(i) 3MnO 4 –+
mol NO
2– 4H+ → 2MnO–3 4 + MnO2 + 2H
–
2 = 2.5  6.07  10 = 1.5175  10
–2 2O 2
conc NaNO 2 = 4  1.5175  10 = 6.07-6.08  10 mol dm
–2 –2 –3 ecf min 2sf
M1 MnO42– as a reactant and MnO4– + MnO2 products identified
2(c)(i) M2 correct equation
3MnO4 + 4H → 2MnO4 + MnO2 + 2H2O
2– + – 2
CH 1: ENERGETICS AND GROUP 2 WORKSHEET 1.1
9701/42 Cambridge International AS & A Level – Mark Scheme May/June 2024
PUBLISHED
4 2024 May/Jun | Paper 42 | Q3
Question Answer Marks

3(a) M1 So = (213.8 + 2  248.2) – (237.8 + 3  205.2) 3

So = –143.2 (J K–1 mol–1)

M2 Ho = (–393.5 + 2  –296.8) – (116.7)

Ho = –1103.8 (kJ mol–1)

M3 Go = Ho – TSo

Go = –1103.8 – (298  –0.1432) = –1061.1 to –1061.4 (kJ mol–1) ecf min 3sf

3(b) M1 Go = Ho – TSo AND Go = 0 2

OR T = Ho / So [1]

M2 T = 261.6 ÷ 0.3655 = 715.7 / 716 / 715 K min 3sf

Question Answer Marks

9701/42 Cambridge International AS & A Level – Mark Scheme February/March 2024
4(a)(i) PUBLISHED
the (3)d and (4)s (sub-shells/orbitals) are close/similar in energy 1
5 2024 Feb/Mar | Paper 42 | Q1
Question Answer Marks
4(a)(ii) 1
1(a)(i) (enthalpy change when) one mole of a substance / solute 1
AND dissolves in water / turns into an aqueous solution
(to form a solution of infinite dilution)

1(a)(ii) there is a (large) increase in entropy OR S is positive OR TS is positive 1

so G is negative / TS outweighs H 1

1(b)(i) anionic charge density decreases (down the group / Cl – to I–) 1

(so hydration enthalpies become less negative / less exothermic because) 1

less attraction of ion to water / the dipole–ion force weakens

1(b)(ii) the difference between Hlatt and Hhyd remains roughly constant 1
OR Hlatt and Hhyd become less exothermic by a similar amount

© Cambridge Hhyd
1(b)(iii)University (K+(g))
Press = –6292024
& Assessment + 21.0 –(–293) = –315 kJ mol–1 Page 7 of 15 1

1(b)(iv) solubility of PbI2 = 7.1  10–9 = x·(2x)2 ∴ x = ∛(¼  7.1  10–9) 1

solubility of PbI2 = 1.21  10–3 mol dm–3 min 2sf 1

1(b)(v) • Pb2+ has a greater charge 2

• Pb2+ is smaller (than K+) OR (Pb2+) smaller ionic radius
• greater attraction between Pb2+ and I–
9701/42 Cambridge International AS & A Level – Mark Scheme February/March 2024
OR ionic bond between Pb2+ and I– is stronger
PUBLISHED
OR lattice energy is more exothermic / more negative
Question mark as  ✓ ✓ Answer Marks
1(c)(i)
2(a)(i) S = 2(155.5) + 2(116.1) – 4(106.3)base
● conjugate – 2(213.6)
of acid I–=205.2
OH– 1
1
9701/42 ● acid II = HNO2 Cambridge
● conjugate International
base of acid II = NO2– AS & A Level – Mark Scheme February/March 2024
S = –514.4 (J K mol ) min 3sf ECF
–1 –1 PUBLISHED 1
2(a)(ii) lone pair on the N can be donated to a proton /H+ 1
Question Answer Marks
OR lone pair on the N can accept / gain a proton /H+
OR lone pair on the N can form a dative bond to a proton/H +
1(c)(ii) G = H – TS AND use of 298 K for T OR clear working to represent this 1
2(a)(iii) H2O + CH3COOH → H3O+ + CH3COO– 1
G =Press
© Cambridge University –203.4 – 298(–514.4
& Assessment 2024 / 1000) Page 5 of 13
Gw )==–50.1 OR–] –50.2 (kJ(K
mol –1) OR G = –50 108.8 J min 3sf ECF from (c)(i) 1
2(b)(i) (K [H+][OH ALLOW w ) = Ka  Kb 1
(negative so spontaneous)
2(b)(ii) endothermic 1
1(c)(iii) increases (in thermal position
AND equilibrium stability down
movesthe group)
right OR as water dissociates more 1
AND (cat)ionic
OR Kw radius / ionwith
increases sizetemperature
increases (down the group)

less
2(b)(iii)
M1 (Kpolarisation of anion / C—O bond / distortion of carbonate ion / CO32–
w increases with temperature so) pH of neutral solution decreases
1
1
OR C–O is less weakened / stronger (down the group)
OR (from graph) Kw = 1.50  10–14 = [H+]2
1(d)(i) ∴ neutral
2H +(aq) +pH2e –= →
–½ H
log (1.50  10–14) = 6.91
2(g) state symbols required 1
OR [H+] = 10–7.00
1(d)(ii) ∴ [OH–] = 1.50  10–14 /21.35
10–7.00 = 1.50  10–7 1
M1 moles of S2O32– = × 0.100 = 2.135  10–3
1000 neutral pH / is alkaline
6 2024 Feb/Mar
M2 pH 7|is Paper
therefore42 above| Q2 1
OR [OH–] > [H+] (so alkaline)
M2 calc of Q = 2.135  10–3  0.5  2  96500 = 206.02 / 205.68 ECF 1
2(c)(i) H2O(l) particles / molecules has more randomness / disorder 1
OR = answer I = 206 / (8  60) = 0.428 OR 0.429 A min 2sf ECF
M3 1
H2O(l) has more ways to arrange particles / energy (than in solid)
1(e)(i) reaction 1 concentrated HNO3 AND concentrated H2SO4 (concentrated seen once) 1
2(c)(ii) H2O(g) particles / molecules has much more randomness / disorder 1
reaction 2 Sn AND (concentrated) HCl
OR 1
H2O(g) has many more ways to arrange particles / energy (than in liquid)
1(e)(ii) step 1 NaNO2 + HCl → HNO2 + NaCl 1
2(c)(iii) +6030 / (70.1 – 48.0) = 272.85 / 272.9 / 273 K (which is 0 °C) 1
step 2 C6H5NH2 + HNO2 + H+ → C6H5N2+ + 2H2O 1
2(d)(i) (+)1.62 V 1
1(e)(iii) nucleophilic (aromatic) substitution 1

© Cambridge University Press & Assessment 2024 Page 7 of 13

© Cambridge University Press & Assessment 2024 Page 6 of 13

ALT ACADEMY PAGE 2 OF 14 A LEVEL CHEMISTRY (CAIE)

 2(d)(iv)
Q is CH3(CH2)4OH 1
AND it is least polar
/ contains a large non-polar hydrocarbon chain
CH 1: ENERGETICS/ stronger id-id forces
AND GROUP 2 with hexane in Q OWTTE WORKSHEET 1.1
[1]

7 2023 Oct/Nov | Paper 41 | Q3

Question Answer Marks
9701/41 Cambridge International AS & A Level – Mark Scheme October/November 2023
3(a)(i) (number of possible) arrangements of particles / energy in a system
PUBLISHED 1
OR
Question measure / degree of disorder in / of a system [1] Answer Marks

3(d)
3(a)(ii) positive
iron(III) /chloride
+ / FeCl 3 1
more gas
AND same statemolecules
/ phase as/ particles
reactantsin/ products
H2O2 [1]
OR more moles / molecules in products / RHS [1]
3(e)(i) hydrogen peroxide / H2O2 AND +2.18 [1] 1
3(b) H = (2  150) – (1  496) (= –196) [1] 1
3(e)(ii) G
OR = –nEocellF [1] 2
H = (2  150) + (4  460) – (1  496) – (4  460) (= –196)
G = (–2  2.18  96500) = –420.7 (kJ mol–1) [1] ecf
3(c) G = H –TS seen or used with correct signs[1] 3
3(f)(i) Nernst: (E =) Eo + (0.059 / z)log[ox] / [red] [1] u / c 2
OR
–238(E==) Eo +–(RT
–196 298S / zF)ln[ox] / [red]
OR
S =(E42=)/ 298
1.82 + (0.059 / 1)log(0.02 / 2)
S = (+)0.141 / 0.1409 (kJ K–1 mol–1)
E = 1.82 + (0.059
OR (+)141 / 1)log(0.02
/ 140.9 (J K–1 /mol–1)
2) = (+)1.702 (V) [1]G
[1] ecf from min 2sf + TS
= H
OR
E
141= 1.82
= (2 + 70)
[(8.314  298 / 1)
+ S(O2(g))  (96
– (2 500) ] ln(0.02 / 2) = (+)1.702
 102)
S, O2(g) = 205 / 204.94 (J K–1 mol–1) [1] ecf
3(f)(ii) H2O2 + 2H+ + 2Co2+ → 2H2O + 2Co3+ [1] 1
ECF for reverse equation from (f)(i) if E > 1.77 V
© UCLES 2023 Page 7 of 18
9701/41
3(g)(i) enthalpy change when one mole Cambridge
of gaseous International
ions AS & A Level – Mark Scheme October/November 2023
1
forms an aqueous solution / dissolves in water [1] PUBLISHED
Question Answer Marks

3(g)(ii) Substances with state symbols: 2

• Al F3(s)
• Al F3(aq) OR Al 3+(aq) + (3)F–(aq)
9701/42 • Al 3+(g) (3)F–(g) Cambridge International AS & A Level – Mark Scheme October/November 2023
[1] PUBLISHED
changes identified and three correct arrow directions (one given) [1]
Question Answer Marks

3(e) 0.64  17  60 = 653 / 652.8 Coulombs [1] 3

652.8 ÷ 1.6x10–19 = 4.08  1021 (number of electrons)
4.08  1021 ÷ 2 = 2.04  1021 (number of atoms Fe) [1]
0.185 ÷ 55.8 = 3.31  10–3 (number of moles Fe atoms
© UCLES 2023 2.04  1021 ÷ 3.31  10–3 = L = 6.153  1023 [1] Page 8 of 18

3(f)(i) S = –179 [1] 3

G = H –TS [1]
–74.7 [1]
9701/42
3(f)(ii) Cambridge
less AND G becomes more positive [1] International AS & A Level – Mark Scheme October/November 2023
1
PUBLISHED
3(g)(iii)
Question Hlatt = –4690 + (3  – 506) – (–209) = –5999 (kJ mol–1) Answer 1
Marks
Question Answer Marks
3(e) 0.64  17  60 = 653 / 652.8 Coulombs [1] 3
4(a)
Question equilibrium
652.8 ÷ 1.6x10constant
–19 for the
= 4.08  10formation
21 of aofcomplex
(number in a solvent
electrons) Answer / Marks1
from its10
4.08 constituents
21
÷ 2 = 2.04[1] 1021 (number of atoms Fe) [1]
4(a)(i) •0.185 Co(OH) OR Co(OH)
 10|–32(H 2O)4 2
8 2023 4(b)Oct/Nov
+2 AND | Paper
÷ 55.8 2= 3.31 42 Q3
(number of moles Fe atoms
 10+213÷3[1]
•2.04 [Co(NH ]2+ OR
)63.31 [Co(NH 3+ 1
 10 –3
= L 3=)6]6.153  1023 [1]
• [CoCl4]2–
4(c) Six [1] 1
3(f)(i) S =two
any –179[1] [1]
all three [2] 3
4(d) G = H –TS
[Fe(EDTA)] –
AND [1] largest K [1]
4(a)(ii) pink
–74.7to[1]
stab
11
blue [1]
4(e) [[CuEDTA] ] ÷ ([[Cu(H2O)6]2+]  [EDTA2–]) = 6.311019 [1]
2– 2
[1]3)2G
3(f)(ii) less 1
4(b)(i) 0.095AND
2Ca(NO
becomes more positive [1]
→ 2CaO + 4NO2 + O2 1
OR Ca(NO3)2 → CaO + 2NO2 + ½O2 [1]
4(f) different E OR different energy gap between d-orbitals [1] 2
Question Answer Marks
© UCLES 2023 absorption of different wavelength OR absorption of different Page 9 frequency
of 18 [1]

4(a) equilibrium constant for the formation of a complex in a solvent / 1

9 2023 Oct/Nov | Paper 42
from its constituents [1] | Q5
Question Answer Marks
9701/42 Cambridge International AS & A Level – Mark Scheme October/November 2023
4(b) +2 AND + 3 [1] PUBLISHED 1
5(a) MgC2O4 → MgO + CO2 + CO [1] 1
4(c)
Question Six [1] Answer Marks1
© UCLES
5(b)2023
4(d) magnesium– ethanedioate
[Fe(EDTA)] AND largest K stab [1]
decomposes PageMg
at lower T because 7 of2+
14has smaller radius than Ca2+ [1] 1
2
so anion is more polarised by Mg2+ [1]
4(e) [[CuEDTA]2–] ÷ ([[Cu(H2O)6]2+]  [EDTA2–]) = 6.311019 [1] 2
5(c) 0.095 ÷ 1000)  0.02 = 5.41  10–5 moles MnO4– [1]
(27.05[1] 3
5.41  10–5  5/2 = 1.3525  10–4 moles C2O42– [1]
4(f) different E OR
0.00338 mol dm–3different
[1] energy gap between d-orbitals [1] 2
absorption of different wavelength OR absorption of different frequency [1]

Question Answer Marks

Question Answer Marks
6(a)(i) donates one lp to metal atom or ion [1] 1
5(a) MgC2O4 → MgO + CO2 + CO [1] 1
6(a)(ii) metal atom or ion bonded to one or more ligands [1] 1

6(a)(iii)
© UCLES 2023 has vacant d-orbitals which are energetically accessiblePage
[1] 7 of 14 1

6(b)(i) octahedral 1
square planar
octahedral [1]

6(b(ii) [Au(dien)(H2O)2Cl]2+ + 2Cl– → [Au(dien)Cl 3] + 2H2O [1] 1

OR [Au(dien)Cl 3] + 2H2O → [Au(dien)(H2O)2Cl]2+ + 2Cl–

6(b)(iii)
ALT ACADEMY PAGE 3 OF 14 A LEVEL CHEMISTRY (CAIE)
 4.0 quartet 1 3

Three correct for one mark, six correct for two marks, nine correct for three marks.
CH 1: ENERGETICS AND GROUP 2 WORKSHEET 1.1
7(e)(ii) one extra peak for NH2 group seen in CDCl3, AND H exchanged for D in D2O 1

10 2023 May/Jun | Paper 41 | Q8

Question Answer Marks

8(a) 1
energy change always always either negative
positive negative or positive

lattice energy 

enthalpy of hydration 

enthalpy of solution 
All correct for one mark

8(b) The energy / enthalpy change when 1 mole of gaseous ions is dissolved in water 1

8(c)(i) M1 use of correct six numbers only 3

682.8 178.2 590 1145 111.9 324.6
9701/41 M2 2 used correctly with Br (2 Cambridge International
111.9 and 2  324.6) AS & A Level – Mark Scheme May/June 2023
M3 correct signs and evaluation to give –2170.6 kJ molPUBLISHED
–1

Question Answer Marks

© UCLES
8(c)(ii)2023 M1 use of correct three numbers only 2170.6 103.1 andPage
157916 of 17 2
M2 correct signs & evaluation –347 kJ mol–1

8(c)(iii) M1 Br– has a smaller ionic radius 2

M2 Br– has stronger attractive forces with water molecules

9701/42 Cambridge International AS & A Level – Mark Scheme May/June 2023

PUBLISHED
11 2023 May/Jun | Paper 42 | Q1
Question Answer Marks

1(a)(i) 1

ARROW 1 starts from O– to the C–O bond

AND
ARROW 2 starts at the C-O bond to other O– ion

1(a)(ii) M1 increases (down the group) 3

M2 (cat)ionic radius / ion size increases (down the group)

OR charge density of M2+ decreases

M3 less polarisation / distortion

of anion / carbonate ion / CO3(2)–

1(b)(i) M1 energy released when one mole of a ionic solid / compound is formed 2
M2 from gas (phase) ion(s) / gaseous ion(s) (under standard conditions)

1(b)(ii)  (Hdecomp / it) becomes more positive/less negative 2

(down the group)

 size /(ionic) radii of oxide ion is smaller (than carbonate ion) ORA
© UCLES 2023 Page 17 of 17
 so Hlatt of oxides becomes ORA
less exothermic faster OR less negative faster
OR changes more OR changes faster

Any two [1], all three [2]

9701/42 Cambridge International AS & A Level – Mark Scheme May/June 2023
1(c)(i) 2MnO4– + 6H+ + 5SO32–  2Mn2+ + 3H2O + 5SO42– PUBLISHED 1
Question Answer Marks

1(c)(ii) M1 M2 3
any two bullets [1] or all four [2]
© UCLES 2023  moles MnO4– = 0.025  22.40 / 1000 = 5.6  10–4 Page 6 of 19
 moles SO32– = 5.6  10–4  5 / 2 = 1.4  10–3 (in 25 cm3) ecf from (c)(i) and bullet 1
 moles SO32– = 1.4  10–2 (in 250 cm3) ecf bullet 2
 mass K2SO3 = 1.4  10–2  158.3 = 2.2162 g ecf bullet 3
OR moles K2SO3 (if 100% pure) = 3.40 ÷ 158.3 = 0.02148

M3
% purity = 100  2.2162 / 3.40 = 65.2 / 65.3 % ecf min 2sf
OR
% purity = 100  0.014 / 0.02148 = 65.2 / 65.3 %

1(d) S() tetrahedral 1

Question Answer Marks

2(a) Any two from: 1

 they have variable / multiple oxidation states OWTTE
 they behave as catalysts
 they form complex ions / complexes
 they form coloured compounds/ions

2(b)(i) (is a molecule or ion formed by a central) metal atom / metal ion bonded / surrounded by (one or more) ligands 1

ALT ACADEMY PAGE 4 OF 14 A LEVEL CHEMISTRY (CAIE)

CH 1: ENERGETICS AND GROUP 2 WORKSHEET 1.1
9701/42 Cambridge International AS & A Level – Mark Scheme May/June 2023
PUBLISHED
12 2023 May/Jun | Paper 42 | Q3
Question Answer Marks

3(a) 1
always always can be either
energy change
positive negative negative or positive

bond energy 

enthalpy change of atomisation 

enthalpy change of formation 

3(b) M1 (enthalpy change) when one mole of gaseous atoms is produced IGNORE energy released 2
M2 from its element(s) in its standard state / standard conditions / 298 K AND 1 atm

3(c) M1 use of correct six numbers only –31 / 285 / 731 / –141 / 798 / 496 3

M2 2  used correctly with Ag (2  285 (570) and 2  731 (1462)) AND 0.5 with O=O (496 (248))

M3 correct signs and evaluation

–31 = (2  285) + (2  731) + (–141) + (798) + x + (0.5  496)

x = –2968 kJ mol–1

3(d)  Ag2Se Ag2S Ag2O OWTTE 2

least exothermic most exothermic

 charge density of anion decreases down the group ORA

/ radius/size of anion increases down the group
/ Se2– largest radius / O2– smallest radius / O has smallest ionic radius

 less attraction between the ions / ionic bond gets weaker (with Ag2Se) ORA

Any two [1], all three [2]

3(e)(i) (Ksp = ) [Ag+]2[SO32–] [1] 1

© UCLES 2023 Page 10 of 19

ALT ACADEMY PAGE 5 OF 14 A LEVEL CHEMISTRY (CAIE)

CH 1: ENERGETICS AND GROUP 2 WORKSHEET 1.1
9701/42 Cambridge International AS & A Level – Mark Scheme February/March 2023
PUBLISHED
13 2023 Feb/Mar | Paper 42 | Q1
Question Answer Marks

1(a)(i) 1
[Ar] ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿⇂

1(a)(ii) 3

one ● for each of the eight species (including state symbol) on the correct line
AND one ● e–

any three [1] any six [2] all nine [3]

1(a)(iii) • EA becomes less negative/ less exothermic (down group / S to Te) 2

• atomic radii increases OR outer shell gets farther from nucleus OR electron added at higher energy level
OR more shielding (of outer shells)

• less nuclear attraction

OR less attraction for incoming/added electron

any two [1] all three [2]

1(a)(iv) M1: O2– (has same charge but) smaller (radius than S2–) ORA 1
OR oxygen has a smaller ion (than S2–)
9701/42 Cambridge International AS & A Level – Mark Scheme February/March 2023
PUBLISHED
M2: stronger ionic bond OR greater attraction between Zn2+ and O2– ORA 1
Question Answer Marks
© UCLES 2023 Page 5 of 17
1(b)(i) S negative AND more moles / molecules of gaseous reactants ORA 1
OR S negative AND moles / molecules of gas are reduced (in the reaction)

1(b)(ii) S = 50.8 + 197.7 – 43.7 – 5.7 = (+)199.1 (J K–1 mol–1) 1

1(b)(iii) G = H – TS ALLOW G = H – TS 1

= +733 – (800 + 273)  0.218 = (+)499.086 (kJ mol–1) min 3sf 1

1(c)(i) Zn(NO3)2 → ZnO + 2NO2 + ½O2 1

OR 2Zn(NO3)2 → 2ZnO + 4NO2 + O2

1(c)(ii) increases (in thermal stability down the group) 1

AND (cat)ion(ic) radius / ion size increases (down the group)

less polarisation / less distortion 1

of anion/ of nitrate ion/NO3– / NO32–
OR less weakening of N—O bond

1(c)(iii) Mg(NO3)2 only ALLOW Mg2+ / magnesium 1

Question Answer Marks

2(a) H3PO2 1

2(b)(i) electrode potential E would become more positive / less negative (than E⦵) 1

lower [H2PO2–] AND shifts equilibrium to the right-hand side 1

2(b)(ii) +1.57 – 0.74 = (+)0.83 (V) 1

© UCLES 2023 Page 6 of 17

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CH 1: ENERGETICS AND GROUP 2 WORKSHEET 1.1
9701/41 Cambridge International AS & A Level – Mark Scheme October/November 2022
PUBLISHED
14 2022 Oct/Nov | Paper 41 | Q1
Question Answer Marks

1(a) 2

M1 K+ (g) and Cl – (g)

AND
KCl (aq) OR K+ (aq) + Cl – (aq)

M2 three correct directional arrows COND M1

1(b) use of data –155, –2493 AND 2  –364 [1] 2

Hhyd Mg2+ = –1920 (kJ mol–1) [1] min 3sf

1(c) • Mg2+ is smaller (than K+) 2

• Mg2+ is greater charge (than K+)
• greater attraction
between Mg2+ and Cl – /between the ions (in MgCl2)
OR stronger ionic bonds (in MgCl2)

1(d)(i) enthalpy change when 1

one mole of gaseous atoms formed from the element (in its standard state at 298 K)

1(d)(ii) enthalpy change when 1

every atom in one mole of gaseous atoms gains one electron
OR one mole of gaseous atoms gains one mole of electrons
9701/41 Cambridge International AS & A Level – Mark Scheme October/November 2022
1(e)(i) PUBLISHED
number of possible arrangements of particles and energy in a system 1
Question Answer Marks

1(e)(ii)
© UCLES 2022
S is positive Page 5 of 16
1
AND KCl(s) → K+(aq) + Cl –(aq) /
ionic lattice solid forms aqueous ions OWTTE [1]
OR
S is positive
AND G is (therefore becomes) negative /
TS is greater than Hsol OWTTE [1]

1(e)(iii) more soluble 1

AND G is more negative at higher T /
TS is more positive at higher T /
–TS is more negative at higher ecf from (e)(ii) [sign S]

Question Answer Marks

9701/42 Cambridge International AS & A Level – Mark Scheme October/November 2022
2(a)(i) PUBLISHED
(homogeneous is in the) same phase / state as reactants 1
15 2022 Oct/Nov | Paper 42 | Q1
Question AND (heterogeneous is in a) different phase / state to reactants
Answer Marks
2(a)(ii) 2– + 2Fe2+ → 2Fe3+ + 2SO 2– [1]
1 S2Ochange) 2
1(a) (energy 8 when one mole of ionic
4 solid is formed from gaseous ions 1
2 2I– + 2Fe3+ → 2Fe2+ + I2 [1]
1(b) (–2237 + 193 + 590 + 1150 + (2  121) – (2  364)) [1] 2
2(a)(iii) reactants are both anions / negatively charged 1
AND
= –790 [1] so they repel each other OWTTE
2(b)(i)
1(c) k[NO]2[O2] OR rate = 8.6  106 [NO]2[O2]
rate =and
–342 1
1
Br atom has larger radius
2(b)(ii) rate = 8.6  106  (7.2  10–4)2  1.9  10–3 1
1(d)(i) energy 8.47 when
rate = change 10–3 (mol
one dm –3 s–1) min 2sf
mole dissolves in water [1] 2
energy change when one mole of gaseous ions dissolves in water [1]
2(c)(i) (reaction is) first order wrt cisplatin / overall 1
1(d)(ii) OR rate– is
(–2237 83directly
+ 1650) proportional
/ 2 [1] to concentration of cisplatin 2
= –335 [1]
2(c)(ii) 0.693 / 2.50  10 = (2.77  10 s)
–5 4 1
1(e)(i) negative
OR In 2 /and
2.50  10–5 = (2.77  104 s) 1
reduction in number of gas molecules
© UCLES 2022 Page 6 of 16
1(e)(ii) TS becomes more negative [1] 2
less feasible AND G becomes positive [1]

Question Answer Marks

2(a)(i) rate = k[NO][O3] 1

2(a)(ii) 1.66  10–8 [1] 2

mol dm–3 s–1 [1]

2(a)(iii) not constant 1

AND
overall second order / not overall first order

2(b)(i) graph is straight line clearly parallel to x-axis 1

2(b)(ii) graph is straight line with negative gradient 1

© UCLES 2022 Page 5 of 11

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CH 1: ENERGETICS AND GROUP 2 WORKSHEET 1.1
9701/41 Cambridge International AS & A Level – Mark Scheme May/June 2022
PUBLISHED
16 2022 May/Jun | Paper 41 | Q1abde
Question Answer Marks

1(a) M1 Hlatt and Hhyd decrease / both become less exothermic / less negative 3

M2 Hlatt decreases / changes less/becomes less exothermic by a smaller extent OR Hhyd decreases / changes
more / dominant factor

M3 Hsol becomes less exothermic / less negative

OR Hsol becomes (more) endothermic / (more) positive
OR Hsol = Hhyd – Hlatt expression AND reaction becomes less exothermic

1(b) Mg: fizzing 1

Ba: (fizzing and) white solid/ppt forms

1(c) M1 solubility of BaSO4 2

= √1.08  10–10 = 1.04  10–5 (mol dm–3)

M2 = 1.04  10–5  233.4 / 10 = 2.43  10–4 (g per 100 cm3 of solution) min 2sf

1(d)(i) –1473 = 180 + 503 + 965 + Hof – 2469 3


Hof of SO42–(g) = –652 kJ mol–1

M1 correct five values used [1]

M2 only correct five values used [1]
M3 correct signs and evaluation [1]
9701/42 Cambridge International AS & A Level – Mark Scheme May/June 2022
1(d)(ii) BaSO4 is more negative/bigger 2
PUBLISHED
 as Ba2+ is smaller OR Ba2+ has a larger charge
Question  stronger force of attraction between the ions Answer Marks
9701/41
2(c) One mark
M1 (in for two correct
a complex Cambridge
/ in the presence International
of ligands) ASare
the d-orbitals & Asplit
Levelinto– two
Marksets
Scheme
of orbitals May/June 2022
3
Two marks for all three correct PUBLISHED
M2 as an electron is excited / promoted
Question Answer Marks

1(e)(i) M3G
M1 visible
o = 0 light
so Tis= absorbed
Hro / So AND colour seen is complementary 2
2(d)(i) Cu(OH)2: (pale / light) blue 1
M2
[Cu(H O)2(NH3)4]2+: dark / deep blue
© UCLES 2022 T = 1322 / 0.616 = 214.3 K Page 5 of 15

T = –58.7 °C min– 2sf

2(d)(ii) (addition of OH ) increases [OH ] AND shifts equilibrium to the right
– 1
1(e)(ii) M1 So = (203 + (70  8) + (2  192)) – (427 – (2  95)) = +530 J K–1 mol–1 3
2(e) 2
M2 Go = Ho – TSo

M3 Go = 133 – (298  0.530) = –24.9 kJ mol–1 ecf 1dp min

M1 one correct 3D diagram with the six correct ligands

Question M2 both 3D diagrams correct Answer Marks

2(a) forms one or more stable ions / compounds / oxidation states 1

17 2022 May/Jun | Paper
with incomplete 42 |filled
/ partially Q3(3)d-orbital(s) / d-shell / d-subshell
Question Answer Marks
2(b)
3(a)(i)  enthalpy change / energy change 12
 one mole of electrons
 (gained by) one mole of gaseous atoms

two for one mark, three for two marks

9701/42 Cambridge International AS & A Level – Mark Scheme May/June 2022
OR
3(a)(ii) (energy required to overcome) the repulsion betweenPUBLISHED
the electron and anion / negative ion 1
2(c)(i)
Question the catalyst and the reactants are in a different state / phase
Answer Marks1

2(c)(ii)
3(a)(iii) M1
 adsorption of reactants
less negative to the surface
/ less exothermic downofthe
thegroup
catalyst 32
M2
 bonds in the
greater the reactants weaken (lowering
distance between the nucleustheand
activation energy)
(the shells of the) electrons
M3 reaction occurs and the
OR atomic radii increases products are desorbed
OR more shielding by inner shells
© UCLES 2022  the less attraction between nucleus Page 7 of 16
and incoming electron (and the less energy released)

two for one mark, three for two marks

© UCLES 2022 Page 6 of 15
3(b) M1 use of correct seven numbers only in calculation / energy cycle 3
M2 only 2  used correctly
M3 correct signs and evaluation ecf

–208 = 131 + 906 + 1733 + 62 + 151 + 2x – 2605

2x = –586
x = –293 kJ mol–1

3(c) first box ticked 1

AND Cd2+ larger / Cd2+ lower charge density
AND less attraction between the ions / weaker ionic bonds

Question Answer Marks

4(a)(i) M1 all five points plotted correctly 2

M2 best-fit straight line (ruler) with negative gradient drawn

4(a)(ii) M1 gradient correctly calculated OR gradient working seen 2

M2 gradient = –So
So evaluated correctly ecf So = (+)160 ± 10 (J K–1 mol–1)

4(b)(i) 2HCO3–  CO32– + CO2 + H2O 1

ALT ACADEMY PAGE 8 OF 14 A LEVEL CHEMISTRY (CAIE)

© UCLES 2022 Page 8 of 16
 x = –293 kJ mol–1

3(c) first box ticked 1

CH 1: ENERGETICSAND
ANDCdGROUP 2 2+ lower charge density
2+ larger / Cd WORKSHEET 1.1
AND less attraction between the ions / weaker ionic bonds

18 2022 May/Jun | Paper 42 | Q4ab

Question Answer Marks

4(a)(i) M1 all five points plotted correctly 2

M2 best-fit straight line (ruler) with negative gradient drawn

4(a)(ii) M1 gradient correctly calculated OR gradient working seen 2

9701/42 M2 gradient = –So Cambridge International AS & A Level – Mark Scheme February/March 2022
So evaluated correctly ecf So = (+)160 ± 10 (J K–1 mol –1)
PUBLISHED
9701/42 Cambridge International AS & A Level – Mark Scheme May/June 2022
Question
4(b)(i) 2HCO3–  CO32– + CO2 + H2O PUBLISHED
Answer Marks1
Question
1(a)(i) oxidising agent [1] Answer Marks2
H2O2 + 2H+ + 2I– → 2H 2O + I2 [1]
4(b)(ii) M1 ionic radius of M+ / cationic radius increases 2
1(a)(ii) OR charge density of ion / M+ decreases down Group 1 2
© UCLES 2022 M1: Kpc (93.8) = [I2(cyclohexane)] ÷ [I2(aq)] Page 8 of 16
93.8 = (0.390 / 15) ÷ (x / 20)
M2 less distortion / polarisation of the anion / HCO3–
ORmass
M2: CO bond / C-Ox/ =
of I2(aq), C=O
5.54less weakened
× 10 –3
(g) ecf
4(c)(i)
1(a)(iii) •M1 Ksolution which resists changes in pH when 22
pc would be lower
•opposes / resistsischange
hexan-2-one in pH(than cyclohexane)
more polar
OR hexan-2-one is polar AND +cyclohexane is non-polar
M2 when small amount of acid / H or alkali / base / OH– is added
19 2022 Feb/Mar
• I2 is |(therefore)
All
Paper 42 | Q1bin hexan-2-one
less soluble
three correct for two marks
4(c)(ii) M1 (with acid) HCO3– + H+ → H2CO3 OR HCO3– + H3O+ → H2CO3 + H2O 2
9701/42 Cambridge International AS & A Level – Mark Scheme February/March 2022
1(b)(i) enthalpy change when one mole of a solute AND dissolves in water to form a solution of infinite dilution 1
M2 (with alkali) H2CO3 + OH– → HCO3– + H2O PUBLISHED
1(b)(ii)
Question –(–629) + (–322) + (–293) = (+)14 (kJ mol–1) Answer Marks13
4(c)(iii) M1 Ka = 10–6.35 = 4.47  10–7
1(b)(iii)
1(e) (cationic)
M1: ([Hg 2+charge
])4.47 density
= 1.00 × –710/ –7decreases
÷ 454.4 Li+ 2.20
to K+×[1] 2
3
=10 –8 ecf 10
–10 (mol dm–3)
M2
so [H+] =energies
lattice  10
become 14.1 = 3.17
less negative / less exothermic AND because less attraction between ions [1]
M2: KpH
sp =
M3 moles= [Hg
–log
2+][I–]2 = 4[Hg2+]3
[H+] = 7.5=ecf = 4.26 × 10
from a×calculated
–29
[H+] ecf
min
1(c)(i) M1: of thiosulfate 0.02230 0.150 = 3.345 × 102sf
–3 2
M3: units = mol3 dm–9 ecf
M2: [Cu2+] = 2 × ½ × 3.345 × 10–3 ÷ 0.0250 = 0.134 (mol dm–3) ecf
Question Answer Marks
20 2022 Feb/Mar
1(c)(ii)
5(a)
starch | Paper 42 | Q2ab 1
Question substance liberated Answer
substance liberated Marks3
1(c)(iii) electrolyte 1
at the anode at the cathode
2(a)(i) 1 mol liquid and 2 mol gas formed from 3 mol solid OR two solid compounds converted to a liquid and a gas 1

2(a)(ii) PbBr
M1: (as2(l)
T increases) T∆S becomesBr2 / bromine
greater (than ∆H) Pb / lead 2
3dxy OR (as T increases) T∆S becomes more positive
concentrated NaCl(aq) Cl2 / chlorine H2 / hydrogen
1(d) M2: (as T increases)
Reduction of Fe3+: feasibility will
2Fe3+ + 2I – →increase
2Fe2+ +asI2 ∆G
[1] becomes more negative 2
Cu(NO3)2(aq)
Regeneration of Fe3+: 2Fe2+ +OS
2 /2O
oxygen
8 (+ H3+2O)
2– → 2Fe + 2SOCu4 / copper
2– [1]
2(b)(i) M1: = 314 + 131 – (19 + 3 × 187) use of values and correct stoichiometry 2
two for one mark, four for two marks, six for three marks
© UCLES 2022 M2: = –135 (J K–1 mol–1) Page 5 of 12

2(b)(ii)
© UCLES 2022 M1: ∆G = 0 ∴ T = ∆H / ∆S = +219.3 × 103 ÷ –(b)(i) Page 9 of 16 2

M2: = 1624(.4) (K)

2(c) M1: 1.00 g Si is 1/28.1 = 0.0356 mol 2

∴ moles of e– needed = 4 × mol Si = 0.142 faraday (3 sf)

M2: Q = It ∴ t = M1 × 96500 ÷ 6 = 2289 (s) ecf

Question Answer Marks

3(a)(i) (d-block) element that forms one or more stable ions with incomplete d subshell / incomplete d orbitals 1

© UCLES 2022 Page 6 of 12

ALT ACADEMY PAGE 9 OF 14 A LEVEL CHEMISTRY (CAIE)

 2(c) all of the (sodium) propanoate (ion) has been 2
protonated / converted to (propanoic) acid / neutralised [1]
CH 1: ENERGETICS AND GROUP 2 WORKSHEET 1.1
H+ is in excess / H+ is 0.1 mol dm–3 (from the H2SO4) [1]

21 2021 Oct/Nov | Paper 41 | Q3

Question Answer Marks

3(a) • enthalpy/energy change 2

• one mole of electrons gained

• by one mole of atoms

• gaseous (atoms)
9701/41 Cambridge International AS & A Level – Mark Scheme October/November 2021
3(b) Ca+(g) → Ca2+(g) + e- [1] PUBLISHED 1
Question Answer Marks

3(c)
© UCLES 2021
M1: selecting correct data 951, 844, 142 only Page 6 of 16
3

M2: evaluation to give 249 (∆Hatom)

OR 2(951) = BE – 2(142) + 2(844)

M3: evaluation to 498 (2 × 249) ecf M2

951 = ∆Hatom –142 + 844

∆Hatom = 249
BE = 498 (kJ mol–1) [3]

3(d)(i) attraction between nucleus / protons / nuclear charge 1

and electron [1]

3(d)(ii) repulsion between 1– ion / electrons of O– 1

and electron [1]

3(e) M1: selecting correct data 951, 1933, 3517 only (ignore signs) 2

M2: evaluation to give –633 (∆Hf) ecf

∆Hf = 951 + 1933 – 3517 = –633 (kJ mol–1) [2]

3(f) ionic charge / charge density (of the ions) [1] 2

greater (attractive) force between the ions [1]

9701/41 Cambridge International AS & A Level – Mark Scheme October/November 2021

PUBLISHED
22 2021 Oct/Nov | Paper 41 | Q4
Question Answer Marks

4(a) • barium carbonate / Ba / BaCO3 2

• larger ionic radius OR smaller charge density of cation / M2+ [1]

© UCLES 2021 Page 7 of 16
• anion / CO32– / carbonate ion is less distorted / less polarised OR C-O / C=O less weakened [1]

4(b) • calcium oxide / calcium hydroxide 3

• CaSO4 / calcium sulfate is more soluble OR BaSO4 is less soluble [1]

• ∆Hlatt and ∆Hhyd are less exothermic / more endothermic (for BaSO4) [1]

• ∆Hhyd is dominant factor / ∆Hhyd change is greater OR ∆Hlatt changes less [1]

4(c)(i) mass of CO2 = 0.02 × 44 = 0.88 g [1] 1

4(c)(ii) (writes correct equation, deduces 3 CO2 per mole) 2

moles of propane = 0.02 / 3 OR 0.00667 OR 1 / 150 [1]

mass of propane = 0.02 / 3 × 44 = 0.293 g [1] ecf M1 × 44 3sf needed

Question Answer Marks

5(a)(i) +2 [1] 1

5(a)(ii) 2

[1]

bond angle labelled 109.5° [1]

© UCLES 2021 Page 8 of 16

ALT ACADEMY PAGE 10 OF 14 A LEVEL CHEMISTRY (CAIE)

CH 1: ENERGETICS AND GROUP 2 WORKSHEET 1.1
9701/42 Cambridge International AS & A Level – Mark Scheme October/November 2021
PUBLISHED
23 2021 Oct/Nov | Paper 42 | Q1
Question Answer Marks

1(a) • enthalpy / energy change / given out / evolved / released 2

• one mole is formed / made [1]

• of compound / solid / lattice / crystal

• (from) gaseous ions [1]

1(b)b S–(g) + e– → S2–(g) [1] 1

1(c)c (555 + 200 – 532 = 223, 223 × 8 = 1784) 3

M1 selecting correct data 555, 200, 532 only, (ignore signs and multipliers) [1]

M2 evaluation to give +223 [1]

M3 multiplying M2 by 8 and evaluation ans (+) 1784 [1]

1(d) (1619 + 555 – 2612 = –438) 2

M1 selecting correct data 1619 555 2612 only, (ignore signs and multipliers) [1]

M2 evaluation to give –438 [1]

1(e)(i) ionic radius / size / sum of ionic radii [1] 2

9701/42 ionic charge / product of ionic charges [1]

Cambridge International AS & A Level – Mark Scheme October/November 2021
PUBLISHED
1(e)(ii) M1 (size tends to make ΔHolatt of radium sulfide) less exothermic since the ions are larger [1] 2
Question Answer Marks
M2 (charge tends to make ΔH latt of radium sulfide) less exothermic since the ions are more highly charged [1]
o

3(e)(ii) states / uses correct Gibbs equation [1] 2

1(e)(iii) (ionic) charge (since) 1
answer = 190 / 191 / 190.0 [1]
AND
3(e)(iii) Becomes less feasible / less spontaneous / 1
ΔHolatt of radium sulfide is more exothermic [1]
AND

© UCLES 2021 because ΔS is negative / TΔS becomes more negative / Page

–TΔS 5 ofbecomes
12 more positive [1]

24 2021 Oct/Nov | Paper 42 | Q4

Question Answer Marks

4(a) strontium nitrate AND because of larger cationic radius [1] 2

NO3– / nitrate ion / anion is less distorted / polarised

OR

N–O or N=O bond less weakened / distorted [1]

4(b) M1 magnesium oxide requires more water to dissolve AND because Sr(OH)2 is more soluble [1] 3

M2 lattice and hydration energies less for Sr(OH)2 [1]

M3 lattice energy is dominant factor / change in lattice energy is greater [1]

4(c)(i) 0.13 [1] 1

4(c)(ii) 26.7 cm3 must be 3SF [1] 1

© UCLES 2021 Page 7 of 12

ALT ACADEMY PAGE 11 OF 14 A LEVEL CHEMISTRY (CAIE)

 3(d) M1: highest [Ag(CN)2]– [Ag(S2O3)2]3– [Ag(NH3)2]+ lowest 2

CH 1: ENERGETICS AND
M2: KstabGROUP
[Ag(CN)2]2– is highest / [Ag(CN)2]– is the most stable WORKSHEET 1.1
OR higher Kstab forms the more stable complex

25 2021 May/Jun | Paper 42 | Q4acd

Question Answer Marks

4(a)(i) M1: energy change when 1 mole of a ionic compound is formed 2

M2: from its gaseous ions under standard conditions

4(a)(ii) ∆Hsol = (–2099) + (2× – 378) – (–2824) 2

∆Hsol = –31 kJ mol–1

M1: use of ×2 as only multiplier

M2: correct signs and evaluation

4(a)(iii) M1: Cu2+ is smaller OR Cu2+ has a higher charge density 2

9701/42 M2: Cu2+attracts water moleculesCambridge International AS & A Level – Mark Scheme
more / stronger May/June 2021
PUBLISHED
OR (Cu2+) forms stronger ion-dipole forces to water molecules
Question Answer Marks
4(b)(i) anode: chlorine / Cl2 1
4(c)(i) cathode: hydrogen / H
measure / degree of (dis)order
2 / randomness (of a system) 1
OR the number of possible arrangements of the particles and their energy (in a given system)
4(b)(ii) M1: Q = 0.75 × 60 × 60 = 2700 C AND 96 500 or 193 000 used 2
4(c)(ii) 1
M2: [a] moles of Ca = 2700 / ∆S
193is000 = 0.0140 ∆S is zero
negative ∆S is positive
[b] mass = 0.0140 × 40.1 = 0.56 g
solid dissolving in water 
© UCLES 2021 Page 8 of 16
water solidifying to ice 

4(c)(iii) 2

9701/42 Cambridge International AS & A Level – Mark Scheme March 2021

PUBLISHED
Question Answer Marks

1(a) Co2+ = [Ar] 3d7 (4s0) 1

Co3+ = [Ar] 3d6 (4s0)

1(b) M1/2: Any two of: 3

• Co3+ is reduced Co2+
two correct for 1 mark, three correct for two marks:
• oxygen gas/O2 is evolved
• starting 3+at +8.6 kJ / in positive region close to the y-axis
• E of Co greater than E of O2
• line passes through x-axis around 100°C
• negative gradient straight– / curve line through the x-axis (no clear positive inflexions)
M3: no change (to [Co(edta)] ) / not feasible OWTTE
4(d)
1(c) Any ∆H
M1: twonegative
of VISUAL , ∆S negative / –
/ – observations: 22
• condensation on tube / steam evolved
M2:
• as temperature
brown gas ∆G
increase,
fumes / brown becomes (more) positive / less negative ora
evolved
OR
• atO2low(er)
formedT,that
(∆Hrelights
more negative
a glowingthan T∆S) so ∆G is negative
splint
OR
• at high(er)
(solid) T, (∆H/ less
dissolves to liquid than T∆S) so ∆G is positive
turnsnegative

1(d)
9701/42 M1: cationic radius / ion size increases (downInternational
Cambridge the group) AS & A Level – Mark Scheme 2
March 2021
M2: less polarisation / distortion of nitrate ion / anion / NO –
PUBLISHED
3

Question Answer Marks

26 2021
© UCLESFeb/Mar
2021
Question
| Paper 42 | Q2a Page 9 of 16
Answer Marks
3(c)(i) M1: E⦵cell for IO3– / H2O2 = –0.68 + 1.19 = +0.51 (∴ feasible) 3
2(a)(i) M2:the
M1 E⦵only cell for H2O2 / Iextracted:
number 2 = +1.77 762,
– 1.19 = +0.58
1560, 496 (∴ feasible) 2
M3:correct
M2 + I2 → 4H2other
5H2O2 multiplier, O + 2IO – + 2H+
four3 numbers used and calculation to the answer
–272 = +416 + ½(496) + 762 + 1560 –141 + 798 + ∆Hlattice
3(c)(ii) 2H∆H
∴ 2O2 → =
lattice
2H 2O + O
–3915 (kJ2 mol–1) ecf 1

3(d)(i)
2(a)(ii) M1:× first
20 order+w.r.t.
[0.9(+2) H2O–2x
0.1(+3)] 2 AND ∴ x ×= 1.5
= 0change in conc. 21 gives increase rate × 1.5 (expts 3 / 4) 13
M2: first order w.r.t. IO3– AND change in conc. × 2 gives increase rate × 2 (as reaction first order w.r.t. H2O2) (expts 1 / 3)
2(a)(iii) M3: zeroth order w.r.t. H + AND change in conc. has no effect on rate (expts 1 / 3 / 4 and 2)
• FeO more exothermic/more negative 2
• Fe2+ has smaller radius/higher charge density (also same charge)
3(d)(ii) 1
•rate greater
= k[H2Oattraction/
2][IO3 ] ecf
–
stronger ionic bonds (between Fe2+ and O2–)
All three for two –5 marks
3(d)(iii) M1: k = 8.82 × 10 ÷ (0.150 × 0.140) = 4.20 × 10–3 min 2sf ecf 2
M2: mol–1 dm3 s–1 ecf

3(e)(i) Ksp = [Pb2+][IO3–]2 1

© UCLES 2021 Page 5 of 13

27 2021 Feb/Mar
3(e)(ii)
M1: 3.69 ×
| Paper
= 42 | Q3f x = ∛(3.69 ×
10–13 x(2x)2 OR 10–13 ÷ 4) 2
M2: = 4.5(2) × 10 (mol dm ) min 2sf ecf
–5 –3

3(f)(i) M1: ∆S = ½(192) + ½(205) + ½(261) + 2(70) – 42 2

M2: (+)427 (J K–1 mol–1) ecf

3(f)(ii) ∆G (always) negative because 1

• ∆H < 0 / negative OR exothermic AND
• ∆S > 0 / positive OR –T∆S < 0 for all T

Question Answer Marks

4(a) (element that forms one or more stable) ions with incomplete/ partially filled 3d-orbitals/d-subshell 1

4(b)(i) Oh AND Td 1

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CH 1: ENERGETICS AND GROUP 2 WORKSHEET 1.1
9701/41 Cambridge International AS & A Level – Mark Scheme October/November 2020
PUBLISHED
28 2020 Oct/Nov | Paper 41 | Q2
Question Answer Marks

2(a)(i) M1 energy released when 1 mole of an ionic compound is formed [1] 2

M2 from gaseous ions (under standard conditions) [1]

2(a)(ii) Ca2+ & O2– have a higher charge / charge density (than Li+ and F–) [1] 1

2(a)(iii) MgO –3600 or more negative AND BaO –3200 or less negative BOTH [1] 1

2(b)(i) BaO(s) + H2O(l) → Ba(OH)2(aq) [1] 1

2(b)(ii) M1 (solubility) increases (down the group) [1] 4

M2 both ∆Hlatt and ∆Hhyd become less exothermic / less negative [1]

M3 ∆Hlatt changes more / is dominant factor [1]

M4 ∆Hsol becomes more negative / more exothermic [1]

2(c) M1: Use of 2 × –348 (EA F) and +158 (bond energy of F2) [1] 3

M2: Use of +147 (at Mg) and +736 and +1450 (IEs of Mg) [1]

M3: evaluation and calculation of their answer

(–1102 – (147 + 158 + 736 + 1450 – 696)) = –2897 (kJ mol–1) [1] ecf

2(d)(i) • (energy change) when an / one electron is added to 2

• each atom / ion in one mole of
• gaseous atoms / ions
mark as    [2]

2(d)(ii) F has greater nuclear charge / more protons 1

AND greater attraction between F atom / nucleus and the electrons
  BOTH [1]

9701/42 Cambridge International AS & A Level – Mark Scheme October/November 2020

© UCLES 2020 PUBLISHED
Page 6 of 15
29 2020 Oct/Nov | Paper 42 | Q3
Question Answer Marks

3(a)(i) (+193 + 242 + 590 + 1150 + (2 × –349)) [1] 2

answer (+)1477 [1]

3(a)(ii) (–795 – 83 – 1477) [1] 2

–2355 [1]

3(a)(iii) (–2355 –(2 × –364)) [1] 2

–1627 [1]

3(a)(iv) Z–Y 1
or
X–W [1]

3(a)(v) less (exothermic) and both ions (in CaCl 2) are larger [1] 1

3(b)(i) soluble barium salt AND soluble sulfate [1] 1

3(b)(ii)
9701/41 less soluble (down the group) [1]Cambridge International AS & A Level – Mark Scheme 4
MAY/JUNE 2020
∆Hlat and ∆Hhyd both decrease down the group [1] PUBLISHED
∆Hhyd decreases more / faster / is dominant factor [1]
Question ∆Hsol gets less exo / more endo [1] Answer Marks

1(d)(ii) [Co(H2O)6]2+ + C6H18N4  [Co(C6H18N4)]2+ + 6H2O 1

OR
[Co(H2O)6]2+ + C6H18N4  [Co(C6H18N4)(H2O)2]2+ + 4H2O

30 2020 May/Jun | Paper 41 | Q2

Question Answer Marks

2(a)(i) M1 solubility increases down the group 4

M2 ∆Hlatt and ∆Hhyd both become less exothermic / less negative

M3 ∆Hlatt changes more (than ∆Hhyd as OH– being smaller than M2+)

M4 ∆Hsol becomes more exothermic / more negative

© UCLES 2020 Page 8 of 14
2(a)(ii) M1 Mg(OH)2 AND Mg2+ has a smaller ionic radii/ Mg2+ has a higher charge density 2

M2 OH- ion is polarised/distorted more

Question Answer Marks

3(a)(i) 6CO2 + 24H+ + 24e-  C6H12O6 + 6H2O 2

ALLOW 6CO2 + 12H+ + 12e-  C6H12O6 + 3O2 for both marks

ALLOW one mark for an unbalanced equation showing the correct species of either equation

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© UCLES 2020 Page 7 of 15
CH 1: ENERGETICS AND GROUP 2 WORKSHEET 1.1
9701/41 Cambridge International AS & A Level – Mark Scheme MAY/JUNE 2020
PUBLISHED
31 2020 May/Jun | Paper 41 | Q7a
Question Answer Marks

7(a)(i) M1 Ksp = [Ag+]2[CO32-] 2

M2 units = mol3 dm-9

7(a)(ii) 1
x = 3√6.3 x 10-12/4 = 1.16 x 10-4 (mol dm-3)
-4
[Ag+] = 1.16 x 10-4 x 2 = 2.33 x 10 (mol dm-3) min 2sf

7(a)(iii) 6.3 x 10-12 = [0.05]2[CO32-] 1

[CO32-] = 2.52 x 10-9 (mol dm-3) min 2sf

7(a)(iv) M1 E=Eo + 0.059log[Ag+] 2

M2 E=0.80 + 0.059log(1.2 x 10-4) = 0.57 V ecf from (a)(ii) min 2sf

7(b)(i) ∆So = 72.7 + 56.5 -96.2 = +33.0 J K-1 mol-1 1

9701/42
7(b)(ii) M1 ∆G = ∆Ho - T∆So Cambridge International AS & A Level – Mark Scheme March 2020 3
PUBLISHED
32 2020 Fe/Mar
M2 ∆G| Paper 42 | Q2a= +55.7 kJ mol-1
= (65.5)- (298x0.033) min 3sf
Question Answer Marks

2(a)(i) M3 ∆G = positive
M1 (thermal so not
stability) feasible/spontaneous
increases (down the group) 3
9701/42 Cambridge International AS & A Level – Mark Scheme
M2 size / radius of metal ion/M2+ increases March 2020
M3 polarisation / distortion of anion / CO32– decreases PUBLISHED
Question Answer Marks
Question
2(a)(ii) M1 lattice energy AND hydration enthalpy become less exothermic Answer Marks3
8(a) M1 a solutionenthalpy
M2 hydration that resists
/ ∆Hchanges
hyd becomesin pHless exothermic more 2
3(b)(i) 2AuCl
M2 3 + 3H
when 2O2 amounts
small → 2Au +of3O 2 + and
acid 6HCl 1
M3 enthalpy change of solution / ∆Hsol alkali
becomesare added to it
less exothermic / more endothermic
3(b)(ii) M1 1st 3 because rate ×2 / doubles when concentration ×2 / doubles 3
8(b)(i)
2(b)(i) K a =
2Al 3+ + order
[NH 3O3][H +w.r.t.
2– +] 3C AuCl
→ 2Al + 3CO 11
M2 First[NH 4 ] H2O2 × 2; AuCl3 × 3 rate × 6 so order = 1 for H2O2
order
+

2(b)(ii) M3 rate
M1 Q = It = k [AuCl 3] [H2O 2]
= 3.5 × 10 × 3 × 602 = 3.78 × 109 C
5 3
8(b)(ii) M1 NH4of
M2 no.
+ + OH-
mol–1e– =  3.78NH × 310+9 /H96500
2O = 3.92 × 104 2
33 2020 Fe/Mar
3(b)(iii) kM3
M2= 1.53
NH| 3Paper
mass ×+Al
10H=3O ×42
+  ×
/ (0.10
27 | 0.50)
NH
3.92 Q3c
×
+ +4=H3.06
4 10 / 32O
= 3.5(3) × 105 g
2
dm mol minute
9 –2 –1

2(b)(iii) 3SiF4 + 2H2O → 2H2SiF6 + SiO2 1

3(c)(i) ∆H4 = (3 ×) electron affinity of fluorine / F 2
∆H6 = (enthalpy change of) formation of AlF3
© UCLES 2020 Page 13 of 15
Question Answer Marks2
3(c)(ii) M1 +326 + 1½ × 158 + 5137 + 3 × –328 + ∆Hlatt = –1504
3(a)(i) M2 ∆H
Mark latt = –6220 (kJ mol )
as • 
–1
2
• voltage of an electrode / half-cell
3(c)(iii) M1 lattice energy of ScF3 should be less exothermic ora 2
•M2 Sc
compared
ion / Sc3+/ connected
larger thantoAl(S)HE / hydrogen
ion / Al half-cell
3+ AND lesser / electrode
attraction between the ions / ionic bonds are weaker
• under standard conditions / 1 mol dm–3, 1 atm, 298 K
3(d)(i) Ksp = [Al3+][F–]3 1

3(d)(ii) Ksp = 6.5 × 10–2 × (3 × 6.5 × 10–2)3 = 4.8 × 10–4 1

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