Chemical energetics

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) ΔH = (–2099) + (2× – 378) – (–2824) 2 sol ΔH = –31 kJ mol–1 sol 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 M2: Cu2+ attracts water molecules more / stronger OR (Cu2+) forms stronger ion-dipole forces to water molecules 4(b)(i) anode: chlorine / Cl 1 2 cathode: hydrogen / H 2 4(b)(ii) M1: Q = 0.75 × 60 × 60 = 2700 C AND 96 500 or 193 000 used 2 M2: [a] moles of Ca = 2700 / 193 000 = 0.0140 [b] mass = 0.0140 × 40.1 = 0.56 g © UCLES 2021 Page 8 of 16 4(c)(i) measure / degree of (dis)order / randomness (of a system) 1 OR the number of possible arrangements of the particles and their energy (in a given system) 4(c)(ii) 1 ΔS is negative ΔS is zero ΔS is positive solid dissolving in water  water solidifying to ice  4(c)(iii) 2 two correct for 1 mark, three correct for two marks: • starting at +8.6 kJ / in positive region close to the y-axis • line passes through x-axis around 100°C • negative gradient straight / curve line through the x-axis (no clear positive inflexions) 4(d) M1: ΔH negative / – , ΔS negative / – 2 M2: as temperature increase, ΔG becomes (more) positive / less negative ora OR at low(er) T, (ΔH more negative than TΔS) so ΔG is negative OR at high(er) T, (ΔH less negative than TΔS) so ΔG is positive © UCLES 2021 Page 9 of 16

3(a) • enthalpy/energy change 2 • one mole of electrons gained • by one mole of atoms • gaseous (atoms) 3(b) Ca+(g) → Ca2+(g) + e- [1] 1 © UCLES 2021 Page 6 of 16 3(c) M1: selecting correct data 951, 844, 142 only 3 M2: evaluation to give 249 (ΔH ) atom OR 2(951) = BE – 2(142) + 2(844) M3: evaluation to 498 (2 × 249) ecf M2 951 = ΔH –142 + 844 atom ΔH = 249 atom 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 (ΔH) ecf f ΔH = 951 + 1933 – 3517 = –633 (kJ mol–1) [2] f 3(f) ionic charge / charge density (of the ions) [1] 2 greater (attractive) force between the ions [1] © UCLES 2021 Page 7 of 16

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 ionic charge / product of ionic charges [1] 1(e)(ii) M1 (size tends to make ΔHo of radium sulfide) less exothermic since the ions are larger [1] 2 latt M2 (charge tends to make ΔHo of radium sulfide) less exothermic since the ions are more highly charged [1] latt 1(e)(iii) (ionic) charge (since) 1 AND ΔHo of radium sulfide is more exothermic [1] latt © UCLES 2021 Page 5 of 12

3(a) • enthalpy/energy change 2 • one mole of electrons gained • by one mole of atoms • gaseous (atoms) 3(b) Ca+(g) → Ca2+(g) + e- [1] 1 © UCLES 2021 Page 6 of 16 3(c) M1: selecting correct data 951, 844, 142 only 3 M2: evaluation to give 249 (ΔH ) atom OR 2(951) = BE – 2(142) + 2(844) M3: evaluation to 498 (2 × 249) ecf M2 951 = ΔH –142 + 844 atom ΔH = 249 atom 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 (ΔH) ecf f ΔH = 951 + 1933 – 3517 = –633 (kJ mol–1) [2] f 3(f) ionic charge / charge density (of the ions) [1] 2 greater (attractive) force between the ions [1] © UCLES 2021 Page 7 of 16

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) M1: (as T increases) TΔS becomes greater (than ΔH) 2 OR (as T increases) TΔS becomes more positive M2: (as T increases) feasibility will increase as ∆G becomes more negative 2(b)(i) M1: = 314 + 131 – (19 + 3 × 187) use of values and correct stoichiometry 2 M2: = –135 (J K–1 mol–1) 2(b)(ii) M1: ∆G = 0 ∴ T = ∆H / ∆S = +219.3 × 103 ÷ –(b)(i) 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

1(a) M1 H and H decrease / both become less exothermic / less negative 3 latt hyd M2 H decreases / changes less/becomes less exothermic by a smaller extent OR H decreases / changes latt hyd more / dominant factor M3 H becomes less exothermic / less negative sol OR H becomes (more) endothermic / (more) positive sol OR H = H – H expression AND reaction becomes less exothermic sol hyd latt 1(b) Mg: fizzing 1 Ba: (fizzing and) white solid/ppt forms 1(c) M1 solubility of BaSO 2 4 = √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 + Ho – 2469 3 f  Ho of SO 2–(g) = –652 kJ mol–1 f 4 M1 correct five values used [1] M2 only correct five values used [1] M3 correct signs and evaluation [1] 1(d)(ii)  BaSO is more negative/bigger 2 4  as Ba2+ is smaller OR Ba2+ has a larger charge  stronger force of attraction between the ions One mark for two correct Two marks for all three correct © UCLES 2022 Page 5 of 15 1(e)(i) M1 Go = 0 so T = Ho / So 2 r M2 T = 132 / 0.616 = 214.3 K T = –58.7 °C min 2sf 1(e)(ii) M1 So = (203 + (70  8) + (2  192)) – (427 – (2  95)) = +530 J K–1 mol–1 3 M2 Go = Ho – TSo M3 Go = 133 – (298  0.530) = –24.9 kJ mol–1 ecf 1dp min Question Answer Marks

3(a)(i)  enthalpy change / energy change 2  one mole of electrons  (gained by) one mole of gaseous atoms two for one mark, three for two marks 3(a)(ii) (energy required to overcome) the repulsion between the electron and anion / negative ion 1 © UCLES 2022 Page 7 of 16 3(a)(iii)  less negative / less exothermic down the group 2  greater the distance between the nucleus and (the shells of the) electrons OR atomic radii increases OR more shielding by inner shells  the less attraction between nucleus and incoming electron (and the less energy released) two for one mark, three for two marks 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

1(a) M1 H and H decrease / both become less exothermic / less negative 3 latt hyd M2 H decreases / changes less/becomes less exothermic by a smaller extent OR H decreases / changes latt hyd more / dominant factor M3 H becomes less exothermic / less negative sol OR H becomes (more) endothermic / (more) positive sol OR H = H – H expression AND reaction becomes less exothermic sol hyd latt 1(b) Mg: fizzing 1 Ba: (fizzing and) white solid/ppt forms 1(c) M1 solubility of BaSO 2 4 = √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 + Ho – 2469 3 f  Ho of SO 2–(g) = –652 kJ mol–1 f 4 M1 correct five values used [1] M2 only correct five values used [1] M3 correct signs and evaluation [1] 1(d)(ii)  BaSO is more negative/bigger 2 4  as Ba2+ is smaller OR Ba2+ has a larger charge  stronger force of attraction between the ions One mark for two correct Two marks for all three correct © UCLES 2022 Page 5 of 15 1(e)(i) M1 Go = 0 so T = Ho / So 2 r M2 T = 132 / 0.616 = 214.3 K T = –58.7 °C min 2sf 1(e)(ii) M1 So = (203 + (70  8) + (2  192)) – (427 – (2  95)) = +530 J K–1 mol–1 3 M2 Go = Ho – TSo M3 Go = 133 – (298  0.530) = –24.9 kJ mol–1 ecf 1dp min 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 H Mg2+ = –1920 (kJ mol–1) [1] min 3sf hyd 1(c) • Mg2+ is smaller (than K+) 2 • Mg2+ is greater charge (than K+) • greater attraction between Mg2+ and Cl – /between the ions (in MgCl ) 2 OR stronger ionic bonds (in MgCl ) 2 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 1(e)(i) number of possible arrangements of particles and energy in a system 1 © UCLES 2022 Page 5 of 16 1(e)(ii) S is positive 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 H OWTTE [1] sol 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

1(a) (energy change) when one mole of ionic solid is formed from gaseous ions 1 1(b) (–2237 + 193 + 590 + 1150 + (2  121) – (2  364)) [1] 2 = –790 [1] 1(c) –342 and 1 Br atom has larger radius 1(d)(i) energy change when one mole dissolves in water [1] 2 energy change when one mole of gaseous ions dissolves in water [1] 1(d)(ii) (–2237 – 83 + 1650) / 2 [1] 2 = –335 [1] 1(e)(i) negative and 1 reduction in number of gas molecules 1(e)(ii) TS becomes more negative [1] 2 less feasible AND G becomes positive [1] 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 H Mg2+ = –1920 (kJ mol–1) [1] min 3sf hyd 1(c) • Mg2+ is smaller (than K+) 2 • Mg2+ is greater charge (than K+) • greater attraction between Mg2+ and Cl – /between the ions (in MgCl ) 2 OR stronger ionic bonds (in MgCl ) 2 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 1(e)(i) number of possible arrangements of particles and energy in a system 1 © UCLES 2022 Page 5 of 16 1(e)(ii) S is positive 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 H OWTTE [1] sol 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

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–) M2: stronger ionic bond OR greater attraction between Zn2+ and O2– ORA 1 © 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(NO ) → ZnO + 2NO + ½O 1 3 2 2 2 OR 2Zn(NO ) → 2ZnO + 4NO + O 3 2 2 2 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/NO – / NO 2– 3 3 OR less weakening of N—O bond 1(c)(iii) Mg(NO ) only ALLOW Mg2+ / magnesium 1 3 2 Question Answer Marks

5(a)(i) palladium, platinum and rhodium / Pt, Pd, Rh 1 5(a)(ii) a catalyst in a different state / phase to the reactants / substrate 1 5(b)(i) measure / degree of disorder / randomness of a system 1 OR the number of possible arrangements of the particles and the energy in a system © UCLES 2023 Page 11 of 17 5(b)(ii) M1 So = (192.8) + 213.8 – 238.2 – 188.8 4 So = -20.4 (J K–1 mol–1) M2 Ho = (–45.9) + (–393.5) – (–101.7) – (–241.8) Ho = -95.9 (kJ mol–1) M3 Go = Ho – TSo M4 Go = –95.9 – (298  –0.0204) = –89.8 (kJ mol–1) 5(c) 4(NH ) CO + 6NO  7N + 8H O + 4CO 1 2 2 2 2 2 2 5(d) 1 5(e)(i) pK = –log K AND pH = –log [H+] 1 a a 5(e)(ii) M1 [H+] = √0.120  2.00  10–4 = 4.89(317)  10–3 2 M2 pH = 2.31 5(e)(iii) % ionisation = 4.89  10-3/0.12  100 = 4.1% 1 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 M2 2 used correctly with Br (2  111.9 and 2  324.6) M3 correct signs and evaluation to give –2170.6 kJ mol–1 © UCLES 2023 Page 16 of 17 8(c)(ii) M1 use of correct three numbers only 2170.6 103.1 and 1579 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 © UCLES 2023 Page 17 of 17

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)  Ag Se Ag S Ag O OWTTE 2 2 2 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 Ag Se) ORA 2 Any two [1], all three [2] 3(e)(i) (K = ) [Ag+]2[SO 2–] [1] 1 sp 3 © UCLES 2023 Page 10 of 19 3(e)(ii) x = 3√1.5  10–14 / 4 = 1.55  10–5 (mol dm-3) 1 [Ag+] = 1.55  10–5  2 = 3.11  10–5 (mol dm-3) min 2sf ecf (e)(i) 3(f)  feasibility / it increases as temperature increases ORA 2  S is positive / S is >0 / entropy change is positive (and H is positive) OR –TS becomes more negative / TS becomes more positive  as G becomes / is negative / G < 0 Any two [1], all three [2] Question Answer Marks

5(a)(i) palladium, platinum and rhodium / Pt, Pd, Rh 1 5(a)(ii) a catalyst in a different state / phase to the reactants / substrate 1 5(b)(i) measure / degree of disorder / randomness of a system 1 OR the number of possible arrangements of the particles and the energy in a system © UCLES 2023 Page 11 of 17 5(b)(ii) M1 So = (192.8) + 213.8 – 238.2 – 188.8 4 So = -20.4 (J K–1 mol–1) M2 Ho = (–45.9) + (–393.5) – (–101.7) – (–241.8) Ho = -95.9 (kJ mol–1) M3 Go = Ho – TSo M4 Go = –95.9 – (298  –0.0204) = –89.8 (kJ mol–1) 5(c) 4(NH ) CO + 6NO  7N + 8H O + 4CO 1 2 2 2 2 2 2 5(d) 1 5(e)(i) pK = –log K AND pH = –log [H+] 1 a a 5(e)(ii) M1 [H+] = √0.120  2.00  10–4 = 4.89(317)  10–3 2 M2 pH = 2.31 5(e)(iii) % ionisation = 4.89  10-3/0.12  100 = 4.1% 1 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 M2 2 used correctly with Br (2  111.9 and 2  324.6) M3 correct signs and evaluation to give –2170.6 kJ mol–1 © UCLES 2023 Page 16 of 17 8(c)(ii) M1 use of correct three numbers only 2170.6 103.1 and 1579 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 © UCLES 2023 Page 17 of 17

3(a)(i) (number of possible) arrangements of particles / energy in a system 1 OR measure / degree of disorder in / of a system [1] 3(a)(ii) positive / + 1 AND more gas molecules / particles in products OR more moles / molecules in products / RHS [1] 3(b) H = (2  150) – (1  496) (= –196) [1] 1 OR H = (2  150) + (4  460) – (1  496) – (4  460) (= –196) 3(c) G = H –TS seen or used with correct signs[1] 3 –238 = –196 – 298S S = 42 / 298 S = (+)0.141 / 0.1409 (kJ K–1 mol–1) OR (+)141 / 140.9 (J K–1 mol–1) [1] ecf from G = H + TS 141 = (2  70) + S(O2(g)) – (2  102) S, O2(g) = 205 / 204.94 (J K–1 mol–1) [1] ecf © UCLES 2023 Page 7 of 18 3(d) iron(III) chloride / FeCl 3 1 AND same state / phase as reactants / H O [1] 2 2 3(e)(i) hydrogen peroxide / H O AND +2.18 [1] 1 2 2 3(e)(ii) G = –nEo F [1] 2 cell G = (–2  2.18  96500) = –420.7 (kJ mol–1) [1] ecf 3(f)(i) Nernst: (E =) Eo + (0.059 / z)log[ox] / [red] [1] u / c 2 OR (E =) Eo + (RT / zF)ln[ox] / [red] OR (E =) 1.82 + (0.059 / 1)log(0.02 / 2) E = 1.82 + (0.059 / 1)log(0.02 / 2) = (+)1.702 (V) [1] min 2sf OR E = 1.82 + [(8.314  298 / 1)  (96 500) ] ln(0.02 / 2) = (+)1.702 3(f)(ii) H O + 2H+ + 2Co2+ → 2H O + 2Co3+ [1] 1 2 2 2 ECF for reverse equation from (f)(i) if E > 1.77 V 3(g)(i) enthalpy change when one mole of gaseous ions 1 forms an aqueous solution / dissolves in water [1] © UCLES 2023 Page 8 of 18 3(g)(ii) Substances with state symbols: 2 • Al F3(s) • Al F3(aq) OR Al 3+(aq) + (3)F–(aq) • Al 3+(g) (3)F–(g) [1] changes identified and three correct arrow directions (one given) [1] 3(g)(iii) H = –4690 + (3  – 506) – (–209) = –5999 (kJ mol–1) 1 latt Question Answer Marks

3(a)(i) (number of possible) arrangements of particles / energy in a system 1 OR measure / degree of disorder in / of a system [1] 3(a)(ii) positive / + 1 AND more gas molecules / particles in products OR more moles / molecules in products / RHS [1] 3(b) H = (2  150) – (1  496) (= –196) [1] 1 OR H = (2  150) + (4  460) – (1  496) – (4  460) (= –196) 3(c) G = H –TS seen or used with correct signs[1] 3 –238 = –196 – 298S S = 42 / 298 S = (+)0.141 / 0.1409 (kJ K–1 mol–1) OR (+)141 / 140.9 (J K–1 mol–1) [1] ecf from G = H + TS 141 = (2  70) + S(O2(g)) – (2  102) S, O2(g) = 205 / 204.94 (J K–1 mol–1) [1] ecf © UCLES 2023 Page 7 of 18 3(d) iron(III) chloride / FeCl 3 1 AND same state / phase as reactants / H O [1] 2 2 3(e)(i) hydrogen peroxide / H O AND +2.18 [1] 1 2 2 3(e)(ii) G = –nEo F [1] 2 cell G = (–2  2.18  96500) = –420.7 (kJ mol–1) [1] ecf 3(f)(i) Nernst: (E =) Eo + (0.059 / z)log[ox] / [red] [1] u / c 2 OR (E =) Eo + (RT / zF)ln[ox] / [red] OR (E =) 1.82 + (0.059 / 1)log(0.02 / 2) E = 1.82 + (0.059 / 1)log(0.02 / 2) = (+)1.702 (V) [1] min 2sf OR E = 1.82 + [(8.314  298 / 1)  (96 500) ] ln(0.02 / 2) = (+)1.702 3(f)(ii) H O + 2H+ + 2Co2+ → 2H O + 2Co3+ [1] 1 2 2 2 ECF for reverse equation from (f)(i) if E > 1.77 V 3(g)(i) enthalpy change when one mole of gaseous ions 1 forms an aqueous solution / dissolves in water [1] © UCLES 2023 Page 8 of 18 3(g)(ii) Substances with state symbols: 2 • Al F3(s) • Al F3(aq) OR Al 3+(aq) + (3)F–(aq) • Al 3+(g) (3)F–(g) [1] changes identified and three correct arrow directions (one given) [1] 3(g)(iii) H = –4690 + (3  – 506) – (–209) = –5999 (kJ mol–1) 1 latt Question Answer Marks

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 H and H remains roughly constant 1 latt hyd OR H and H become less exothermic by a similar amount latt hyd 1(b)(iii) H (K+(g)) = –629 + 21.0 –(–293) = –315 kJ mol–1 1 hyd 1(b)(iv) solubility of PbI = 7.1  10–9 = x·(2x)2 ∴ x = ∛(¼  7.1  10–9) 1 2 solubility of PbI = 1.21  10–3 mol dm–3 min 2sf 1 2 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– OR ionic bond between Pb2+ and I– is stronger OR lattice energy is more exothermic / more negative mark as  ✓ ✓ 1(c)(i) S = 2(155.5) + 2(116.1) – 4(106.3) – 2(213.6) – 205.2 1 S = –514.4 (J K–1 mol–1) min 3sf ECF 1 © Cambridge University Press & Assessment 2024 Page 5 of 13 1(c)(ii) G = H – TS AND use of 298 K for T OR clear working to represent this 1 G = –203.4 – 298(–514.4 / 1000) G = –50.1 OR –50.2 (kJ mol–1) OR G = –50108.8 J min 3sf ECF from (c)(i) 1 (negative so spontaneous) 1(c)(iii) increases (in thermal stability down the group) 1 AND (cat)ionic radius / ion size increases (down the group) less polarisation of anion / C—O bond / distortion of carbonate ion / CO 2– 1 3 OR C–O is less weakened / stronger (down the group) 1(d)(i) 2H+(aq) + 2e– → H (g) state symbols required 1 2 1(d)(ii) 21.35 1 M1 moles of S 2 O 3 2– = × 0.100 = 2.135  10–3 1000 M2 calc of Q = 2.135  10–3  0.5  2  96500 = 206.02 / 205.68 ECF 1 M3 = answer I = 206 / (8  60) = 0.428 OR 0.429 A min 2sf ECF 1 1(e)(i) reaction 1 concentrated HNO AND concentrated H SO (concentrated seen once) 1 3 2 4 reaction 2 Sn AND (concentrated) HCl 1 1(e)(ii) step 1 NaNO 2 + HCl → HNO 2 + NaCl 1 step 2 C H NH + HNO + H+ → C H N + + 2H O 1 6 5 2 2 6 5 2 2 1(e)(iii) nucleophilic (aromatic) substitution 1 © Cambridge University Press & Assessment 2024 Page 6 of 13

4(a) M1 voltage of a half-cell compared to SHE 2 M2 1 mol dm–3 AND 101 kPa AND 298 K 4(b)(i) 3 M1 Pt Ag Ag+ M2 gas delivery system H H+ 2 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 4(c) enthalpy change when one mole of a solute dissolves in water 1 4(d)(i) 2 M1 two arrows in blue M2 correct species in red with all state symbols © Cambridge University Press & Assessment 2024 Page 8 of 18 4(d)(ii) H = –811.6 (kJ mol–1) 1 latt 4(e) M1 Mg(NO ) (s), NaNO (s), RbNO (s) 3 3 2 3 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

1(a) M1 magnesium > calcium > strontium 4 M2 H and H both become less exothermic / less negative latt hyd M3 H changes less OR H is dominant factor latt hyd M4 H becomes less exothermic / less negative / more positive / more endothermic sol 1(b) M1H / energy change when 1 mole of an ionic solid / compound is formed 2 M2 from gaseous ions (under standard conditions) 1(c) M1 as ionic radii increases AND H less exothermic 2 latt M2 as ionic charge increases AND H increases/more exothermic latt 1(d)(i) 3 any two [1] any three [2] all four [3] © Cambridge University Press & Assessment 2024 Page 5 of 15 1(d)(ii) M1 selection of ONLY six correct values (–381, 89, 419, 279, –200, 640) AND use of  2 as only multiplier with K 2 M2 correct evaluation of data used ecf –381 = (89  2) + (419  2) + 279 + (–200) + 640 + Ho latt Ho = –2116 (kJ mol–1) latt 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

4(a) M1 voltage of a half-cell compared to SHE 2 M2 1 mol dm–3 AND 101 kPa AND 298 K 4(b)(i) 3 M1 Pt Ag Ag+ M2 gas delivery system H H+ 2 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 4(c) enthalpy change when one mole of a solute dissolves in water 1 4(d)(i) 2 M1 two arrows in blue M2 correct species in red with all state symbols © Cambridge University Press & Assessment 2024 Page 8 of 18 4(d)(ii) H = –811.6 (kJ mol–1) 1 latt 4(e) M1 Mg(NO ) (s), NaNO (s), RbNO (s) 3 3 2 3 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

2(a) becomes less negative/less exothermic (down the group / from F to I) 2 AND due to increase in (ionic) radius / size [1] decreased attraction to water OR weaker ion-dipole force to water [1] 2(b)(i) CaF (aq) OR Ca2+ (aq) + 2F– (aq) [1] 1 2 2(b)(ii) • atomisation energy of Ca / H 2 at(o)m)) • atomisation energy of F / H (2) at(o)m)) OR F-F bond energy / BE of F-F • first ionisation energy / IE of Ca / H 1 i1 • second ionisation energy / IE of Ca / H 2 i2 • (first) electron affinity /EA of F / H ea any three for [1] ALL five [2] © Cambridge University Press & Assessment 2024 Page 6 of 14 2(b)(iii) energy change / energy released 2 when one mole of an ionic compound is formed [1] from its gaseous ion(s) [1] 2(b)(iv) (H = ) (change) 3 – (change) 1 [1] 1 latt OR (H = ) H(CaF (s)) – H(Ca2+(g)) – 2H(F–(g)) latt f 2 f f 2(c) expression involves four correct numbers (13, 1395, 1214, 1650) 2 AND 2 times [1] 1395 – 1650 + 2x = –1214+13 H(F–(g))  = –473 [1] ecf f correct answer [2] 2(d) number of possible arrangements of particles / energy in a system [1] 1 2(e) states or clearly uses G = H - TS [1] 2 OR 6000 = 30 000 – (298  S) S = (+)80.5(4) [1] min 3sf ecf correct answer [2] 2(f) becomes more soluble 1 AND S is positive / TS is positive / –TS is negative (as T inc) [1] (so G becomes more negative) © Cambridge University Press & Assessment 2024 Page 7 of 14

2(a) H increases from left to right due to increase in charge [1] 3 hyd ionic radius decreases from left to right [1] causing increased attractive force to water molecules [1] 2(b)(i) Mg2+ (aq) + 2Cl– (aq) [1] 1 2(b)(ii) Enthalpy change of hydration of magnesium ions and chloride ions [1] 2 H Mg2+ + 2H Cl– [1] hyd hyd 2(b)(iii) Selects 155, 1920 and 364 only [1] 3 2  364 [1] answer –2493 [1] 2(c) The number of arrangements of the particles and of the energy in the system [1] 1 2(d) G = H – TS [1] 2 answer +10.5 [1] 2(e)(i) No, G is positive [1] 1 2(e)(ii) Becomes more / soluble because G becomes more negative / less positive / smaller / closer to zero [1] 1 Question Answer Marks

2(a) becomes less negative/less exothermic (down the group / from F to I) 2 AND due to increase in (ionic) radius / size [1] decreased attraction to water OR weaker ion-dipole force to water [1] 2(b)(i) CaF (aq) OR Ca2+ (aq) + 2F– (aq) [1] 1 2 2(b)(ii) • atomisation energy of Ca / H 2 at(o)m)) • atomisation energy of F / H (2) at(o)m)) OR F-F bond energy / BE of F-F • first ionisation energy / IE of Ca / H 1 i1 • second ionisation energy / IE of Ca / H 2 i2 • (first) electron affinity /EA of F / H ea any three for [1] ALL five [2] © Cambridge University Press & Assessment 2024 Page 6 of 14 2(b)(iii) energy change / energy released 2 when one mole of an ionic compound is formed [1] from its gaseous ion(s) [1] 2(b)(iv) (H = ) (change) 3 – (change) 1 [1] 1 latt OR (H = ) H(CaF (s)) – H(Ca2+(g)) – 2H(F–(g)) latt f 2 f f 2(c) expression involves four correct numbers (13, 1395, 1214, 1650) 2 AND 2 times [1] 1395 – 1650 + 2x = –1214+13 H(F–(g))  = –473 [1] ecf f correct answer [2] 2(d) number of possible arrangements of particles / energy in a system [1] 1 2(e) states or clearly uses G = H - TS [1] 2 OR 6000 = 30 000 – (298  S) S = (+)80.5(4) [1] min 3sf ecf correct answer [2] 2(f) becomes more soluble 1 AND S is positive / TS is positive / –TS is negative (as T inc) [1] (so G becomes more negative) © Cambridge University Press & Assessment 2024 Page 7 of 14

1(a) (Ag(2+) is a) forms stable ion(s) 1 with incomplete d orbitals / an incomplete d subshell 1(b)(i) Ag(s) → Ag(g) 2 Ag(s) + ½F (g) → AgF(s) 2 1(b)(ii) H/energy change 2 when 1 mole of an ionic solid / lattice / crystal / compound is formed from gas phase ions / gaseous ion(s) (under standard conditions) 1(b)(iii) M1: 5 numbers and no multipliers 2 –203 = +289 + 732 + 79 + EA + (–955) 1 M2 correct sign and evaluation ∴ EA = –348 (kJ mol–1) ecf min 3sf 1 1(c)(i) enthalpy change of solution of AgF = –(–955) + (–464) + (–506) 1 = –15 (kJ mol–1) 1(c)(ii) AgF is (slightly) soluble / yes 1 AND (the enthalpy change of solution is slightly) exothermic / negative ecf from Q1(c)(i) 1(d)(i) greater the distance between the nucleus and (the shells of the) electrons 2 OR atomic radii increases / atomic size increases / more shells OR more shielding by inner shells the less attraction between nucleus / protons AND incoming electron / added electron 1(d)(ii) halide/X–/ions get larger (down the group) 1 AND decreasing attraction between ions / weaker ionic bond 1(e)(i) E⦵cell = +0.342 – (+0.222) = (+)0.120 (V) min 2sf 1 Cu2+ + 2Ag + 2Cl– → Cu + 2AgCl 1 © Cambridge University Press & Assessment 2025 Page 5 of 14 1(e)(ii) M1: [Cl–(aq)] = 10  36.0 / 74.6 = 4.83 (4.8257) mol dm–3 3 0.059 1 M2: E = 0.222 + log ecf M1 1 M1 M3: E = (+)0.182 (V) min 2sf ecf M1 Question Answer Marks

4(a)(i) energy change when one mole of gaseous ions dissolves in water 1 4(a)(ii) M1: K+ has larger radius AND K+ has smaller charge 2 M2: Ca2+ has larger attraction for H O 2 4(a)(iii) energy change when one mole of ionic solid forms from gaseous ions 1 4(a)(iv) M1: –1650–(2  506) + 2602 2 M2: –60 4(b) M1: G = H –TS AND T = 298 2 M2: –174.5 [1] Question Answer Marks

3(a)(i) number of possible arrangements of particles/molecules AND energy in a system 1 3(b)(i) T melting OR solid to liquid 1 1 T boiling OR liquid to gas 2 3(b)(ii) change in disorder from a liquid to a gas is much bigger than solid to liquid 1 OR change in intermolecular distance between liquid & gas is much bigger than solid to liquid 3(c) M1: ΔHo = (3  –393.5) – ((–824.2) + (3  –110.5)) = –24.8 (kJ mol–1) 3 r M2: ΔGo = ΔHo – TΔSo AND T= 723 used M3: (24800 – 36200) ÷ 723 = – [3  213.8 + 2  27.3) – (87.4) – (3x)] –15.8 = –608.6 + 3x, so 3x = 592.8, so x = 197.6 / 198 (J K–1 mol–1) min 3sf 3(d) • (as temperature increases) the reaction is less feasible 2 • as ΔGo becomes less negative / (more) positive • due to TS becoming more negative / –TS (becoming more) positive M1 / M2: any two of the above bullets [1] all three [2] Question Answer Marks

4(a)(i) energy change when one mole of gaseous ions dissolves in water 1 4(a)(ii) M1: K+ has larger radius AND K+ has smaller charge 2 M2: Ca2+ has larger attraction for H O 2 4(a)(iii) energy change when one mole of ionic solid forms from gaseous ions 1 4(a)(iv) M1: –1650–(2  506) + 2602 2 M2: –60 4(b) M1: G = H –TS AND T = 298 2 M2: –174.5 [1] Question Answer Marks

2(a)(i) white / steamy fumes 1 2(a)(ii) number of (possible) arrangements of particles AND energy in a system 1 2(a)(iii) due to the large number increase in gas molecules formed (in the reaction) 1 OR there are much more gaseous molecules in the products OR six gas moles are produced 2(a)(iv) M1: H = (–859) + (2 x –297) + (4 x –92) – (–1460) – (2 x –246) OR +131 kJ mol–1 3 r M2: use of G = H – TS AND use of 298 (or 273+25) for T M3: G = 131 – (298 x 0.768) = –97.9 (kJ mol–1) 2(b)(i) 2Ba2+(aq) + Cr O 2-(aq) + H O(l) ⎯⎯→ 2BaCrO (s) + 2H+(aq) 1 2 7 2 4 2(b)(ii) M1: d orbital(s) of two different energies / d-d splitting occurs 3 OR d orbital(s) / d (sub)-shell splits / d-d gap OR (inferred from movement of an electron) from lower d to higher d orbital M2: electron(s) promoted / excited OR electron(s) moves to higher (d–) orbital OR electron(s) jumps up (to d– orbital) / jumps to higher (d–orbital) M3: wavelength / frequency / light / photon / h / hf absorbed OR radiation / energy from visible (region) absorbed AND colour (seen) is complementary OR wavelength / frequency / colour / light not absorbed is transmitted / reflected / seen 2(c) M1: H and H decrease / both become less exothermic / less negative 3 latt hyd M2: H changes more / dominant factor / changes faster hyd OR H changes less / becomes less exothermic by a smaller extent latt M3: H becomes less exothermic / less negative sol OR H becomes (more) endothermic / (more) positive sol © Cambridge University Press & Assessment 2025 Page 9 of 17

8(a)(i) M1: energy change / released when 1 mole of a (ionic) solid / lattice / crystal / compound is formed 2 M2: from gas (phase) ions / gaseous ions (under standard conditions) 8(a)(ii) (enthalpy change when) 1 mole of a substance / solid / solute / molecule AND dissolves in water (to give a solution of infinite 1 dilution) 8(b) H (NaCl) = H (Na+) + H (Cl–) – H (NaCl) 1 sol hyd hyd latt © Cambridge University Press & Assessment 2025 Page 16 of 17 8(c) 3 M1 / M2: any two [1] all four [2] M3: all state symbols for the formula are present and correct 8(d) M1: (more exothermic because) Cl– OR NO – because its (ionic) radius / size is smaller 2 3 M2: (more exothermic because) (ion-dipole) attraction / bond between it and water is stronger OR M1: NO – because it has lone pairs on O / more lone pairs 3 M2: which can form hydrogen bonds with water © Cambridge University Press & Assessment 2025 Page 17 of 17

4(a) energy required when one mole of gaseous atoms is formed from the element 1 4(b) energy released when one mole of gaseous atoms gains one mole of electrons and becomes one mole of gaseous 1– ions 1 4(c) M1: iodine atom has greater radius 2 M2: less attraction between nucleus and incoming electron in iodine 4(d) M1: ½ (–486 + (2  364) ) 2 M2: +121 Question Answer Marks

2(a) M1: (solubility) decreases down the group 4 M2: H and H decrease / become less exothermic / less negative latt hyd M3: H decreases / changes more / dominant factor / becomes less exothermic by a larger extent hyd OR H decreases / changes less / changes slower latt M4: H becomes less exothermic / less negative OR H becomes (more) endothermic / (more) positive sol sol 2(b)(i) Ag CrO (s) ⇌ 2Ag+(aq) + CrO 2–(aq) 1 2 4 4 2(b)(ii) M1: K = [Ag+]2 [CrO 2-] 3 sp 4 M2: 3√(1.12×10−12÷4) = 6.54  10–5 M3: [Ag+] = 2  6.54  10–5 = 1.31  10–4 (mol dm–3) min 2sf © Cambridge University Press & Assessment 2025 Page 8 of 16 2(c)(i) K = 10–6.49 = 3.236  10–7 2 a [H+]2 = 8.08984  10–9 M1: [H+] = 8.99  10–5 M2: pH = –log(8.99  10–5) = 4.05 min 2sf 2(c)(ii) conjugate acid H CrO 1 2 4 AND conjugate base CrO 2– 4 2(d)(i) (the energy change / released when) one mole of gaseous atoms become 1– ions / gain one mole of electrons 1 2(d)(ii) due to repulsion between negative ion and incoming / gained electron 1 2(d)(iii) 2Ag+(g) S2–(g) → Ag S(s) 1 2 2(d)(iv) M1: selection of correct six numbers only: –33, 731, 279, –200, 532, –2677 3 M2: use of  2 as only multiplier with Ag M3: correct signs and evaluation of data –33 = 2H + (2  731) + 279 + (–200) + 532 + (–2677) at 2H = 571 (kJ mol –1) at H = +285.5 (kJ mol –1) at 2(e) Ag S smaller / less negative / less exothermic lattice neergy AND larger ionic radius (of cation) 1 2 © Cambridge University Press & Assessment 2025 Page 9 of 16

3(a) number of possible arrangements of particles and energy in a system 1 3(b)(i) 1 process S is negative S is positive steam condensing into water ✓ solid KCl dissolving into water ✓ 3(b)(ii) M1: three numbers and correct multipliers used 2 M2: correct signs and evaluation So = (3  203.0) + (223.1) – (2  281.6) = (+)268.9 (J K–1 mol–1) 3(c)(i) positive 1 AND gas produced / on the right OR less / no gas on left 3(c)(ii) M1 and M2: calculation of gradient from the graph for S 4 gradient = (–)140 / 900 = (–)0.1556 So = (–)0.1556  1000 = (+)155.6 ± 5 (J K–1 mol–1) M3: T = 1120 ± 5 (K) M4: Ho = (+)172 ± 10 (kJ mol–1) © Cambridge University Press & Assessment 2025 Page 10 of 16

1(a)(i) (d-block element which forms one or more) stable ions/compounds with an incomplete d orbital(s)/d sub-shell 1 1(a)(ii) 1 3d 4s 4p [Ar] ↿ ↿ ↿ ↿ ↿ ↿ 1(a)(iii) 1 1(a)(iv) more than one (stable) oxidation state / exist in variable oxidation states 2 vacant/empty (d) orbitals are energetically accessible OR vacant/empty (d) orbitals can form dative bonds with ligands OR vacant/empty (d) orbitals can accept lone pair to bond with ligands 1(b)(i) M1: emf / potential / voltage of an electrode / half-cell compared / connected to SHE / hydrogen half-cell 2 M2: 1 mol dm–3, 1 atm, 298 K © Cambridge University Press & Assessment 2026 Page 7 of 18 1(b)(ii) M1: (Cr2+ ions) can be oxidised to Cr3+ 2 M2: by O in acidic conditions / O with H+ because E⦵ =(+)1.64 (V) 2 2 cell OR by O in acidic conditions / O with H+ as E⦵ O / H+ is more positive than Cr3+ / Cr2+ 2 2 2 OR by O in alkaline conditions / O with OH– because E⦵ = (+) 1.66 (V) OR +0.81 V 2 2 cell OR by O in alkaline conditions / O with OH– as E⦵ O / OH– is more positive than Cr3+ / Cr2+ 2 2 2 1(b)(iii) for green solid 1 Cr3+(aq) + 3OH–(aq) → Cr(OH) (s) 3 1(b)(iv) for yellow solution 2 4Cr(OH) (s) + 8OH–(aq) + 3O (g) → 4CrO 2–(aq) + 10H O(l) 3 2 4 2 correct species [1] correct balancing [1] 1(c)(i) gaseous products are formed 1 OR increase in moles / molecules of gas 1(c)(ii) 915 = 192 + 81 + 4S – 114 2 ∴ S = ¼  756 = (+) 189 (J K–1 mol–1) 1(c)(iii) G always / needs / is to be negative / <0 because H < 0 / negative / exothermic S > 0 / positive / –TS is negative 2 high activation energy / E prevents reaction a 1(d)(i) (enthalpy change/energy released when) 2 • 1 mole of • gaseous atoms accept • one mole of e– any two [1], all three [2] © Cambridge University Press & Assessment 2026 Page 8 of 18 1(d)(ii) M1: (2  398 + 3248 + 2  5233 + 3  702) 2 M2: –1128 – 14112 = –15240 (kJ mol–1) min 4 sf 1(d)(iii) ● Cr O less exothermic / less negative 2 2 3 ● Cr3+ (has same charge but) larger radius / is less charge dense ● less attraction between Cr3+ and O2– / the different ions OR weaker bond between Cr3+ and O2– any two [1], all three [2] © Cambridge University Press & Assessment 2026 Page 9 of 18

1(a) Ca2+(g) + S2–(g) → CaS(s) 2 If reverse lattice energy equation is given M1 all species CaS(s) → Ca2+(g) + S2–(g) award [1] if all M2 all state symbols species and state symbols are correct. 1(b) (ionic) bonds are formed / the ions bond 1 A attractive force  distance. A electrostatic attraction  distance. I electrostatic attraction alone. I exothermic / energy released. I ions come together. I formation of the lattice / solid / CaS. I reference to energy required to break bonds. CON energy required to form bonds. CON bond formed is clearly covalent, e.g. Ca–S. 1(c) M1 MgO CaS KCl 3 A for all four charges M2 CaS (and MgO) the ions / particles are more highly charged than KCl I nuclear charge or KCl has ions with the smallest charge ora I charge density for M2 and M3 M3 MgO the ions/particles are smaller than CaS or MgO has the smallest ions ora 1(d)(i) (enthalpy of) atomisation (of Cl) or bond energy (Cl–Cl ) and 1 If this is given in symbols we will accept H , H or H . If symbols and at atom atomisation words are given, I the symbols. (first) electron affinity (of chlorine) If this is given in symbols we will accept H , H , H or EA EA1 electron affinity H . If symbols and words are first electron affinity given, I the symbols. CON electron affinity of chloride / Cl – / Cl 2 © Cambridge University Press & Assessment 2026 Page 7 of 21 Trace ID: PC-F9EA0E463D Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P97a01p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/41 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 1(d)(ii) M1 chlorine value doubled / –486 seen / 2x – 243 seen 2 ecf use of 1x – 243 for ans = –400 M2 answer –643 (working needed). 2336 +(2x – 243) –2493 = –643 –643 scores [2] 1(e) M1 Correct G = H – TS and use of T = 873 3 CON G for G. A  . Gibbs M2 G = –36 / –35.8 / –35.78 / –35.775 ecf T only If T = 600 then [1] ecf for +12. CON G for G but do not penalise twice M3 it is (feasible) and If in words we need ‘change’. G is negative / G is less than zero UNC (Note: G now in bold) I G is exothermic. M3 is UNConditional on the basis it is a correct answer on the answer line. If A G in words ‘Gibbs Free Energy the calculation has not been worked through to an answer, or has given a change’ negative value for G, there is no M3 for ‘not feasible because G is positive’. 1(f) • higher 2 ⚫✓✓ any two for one mark. All three for two marks. • Ca2+ larger (than Mg2+) ora or ionic radius / size I atomic • (In CaCO ) less distortion / polarisation of anion / CO y– ora or group 2 trend but up / down the group 3 x must be clear. all standalone I references to ionic bond strength. ora for MgCO 3 but must be stated or Ca2+ less polarising on anion / CO x y– I ‘depolarisation’. or C-O / C=O bond (must be identified, e.g. C–O) is less weakened by Ca2+. © Cambridge University Press & Assessment 2026 Page 8 of 21 Trace ID: PC-F9EA0E463D Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P97a01p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/41 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

1(a) Ca2+(g) + S2–(g) → CaS(s) 2 If reverse lattice energy equation is given M1 all species CaS(s) → Ca2+(g) + S2–(g) award [1] if all M2 all state symbols species and state symbols are correct. 1(b) (ionic) bonds are formed / the ions bond 1 A attractive force  distance. A electrostatic attraction  distance. I electrostatic attraction alone. I exothermic / energy released. I ions come together. I formation of the lattice / solid / CaS. I reference to energy required to break bonds. CON energy required to form bonds. CON bond formed is clearly covalent, e.g. Ca–S. 1(c) M1 MgO CaS KCl 3 A for all four charges M2 CaS (and MgO) the ions / particles are more highly charged than KCl I nuclear charge or KCl has ions with the smallest charge ora I charge density for M2 and M3 M3 MgO the ions/particles are smaller than CaS or MgO has the smallest ions ora 1(d)(i) (enthalpy of) atomisation (of Cl) or bond energy (Cl–Cl ) and 1 If this is given in symbols we will accept H , H or H . If symbols and at atom atomisation words are given, I the symbols. (first) electron affinity (of chlorine) If this is given in symbols we will accept H , H , H or EA EA1 electron affinity H . If symbols and words are first electron affinity given, I the symbols. CON electron affinity of chloride / Cl – / Cl 2 © Cambridge University Press & Assessment 2026 Page 7 of 21 Trace ID: PC-F2CDA313A1 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P97a01p/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/43 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 1(d)(ii) M1 chlorine value doubled / –486 seen / 2x – 243 seen 2 ecf use of 1x – 243 for ans = –400 M2 answer –643 (working needed). 2336 +(2x – 243) –2493 = –643 –643 scores [2] 1(e) M1 Correct G = H – TS and use of T = 873 3 CON G for G. A  . Gibbs M2 G = –36 / –35.8 / –35.78 / –35.775 ecf T only If T = 600 then [1] ecf for +12. CON G for G but do not penalise twice M3 it is (feasible) and If in words we need ‘change’. G is negative / G is less than zero UNC (Note: G now in bold) I G is exothermic. M3 is UNConditional on the basis it is a correct answer on the answer line. If A G in words ‘Gibbs Free Energy the calculation has not been worked through to an answer, or has given a change’ negative value for G, there is no M3 for ‘not feasible because G is positive’. 1(f) • higher 2 ⚫✓✓ any two for one mark. All three for two marks. • Ca2+ larger (than Mg2+) ora or ionic radius / size I atomic • (In CaCO ) less distortion / polarisation of anion / CO y– ora or group 2 trend but up / down the group 3 x must be clear. all standalone I references to ionic bond strength. ora for MgCO 3 but must be stated or Ca2+ less polarising on anion / CO x y– I ‘depolarisation’. or C-O / C=O bond (must be identified, e.g. C–O) is less weakened by Ca2+. © Cambridge University Press & Assessment 2026 Page 8 of 21 Trace ID: PC-F2CDA313A1 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P97a01p/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/43 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

4(a)(i) one mole of gaseous atoms and 1 A one e– is added to each atom in 1 mole gains one mole of electrons / forms one mole of gaseous 1– ions of gaseous atoms. CON energy required/needed. I equations. CON incorrect near misses, e.g. ‘1 mol of e– is added to each atom…’ or ‘1 e– is added to 1 mol of gaseous atom…’. 4(a)(ii) (first electron affinity is) 2 • less exothermic / less –ve down group OWTTE A decreases / less exothermic. I less energy needed. • (atomic) radius / size increasing or more shells or I charge density. more shielding CON ionic radius for second bullet. • less attraction of nucleus for valence / incoming / outer shell electron A nuclear charge / protons for nucleus. ✓✓ ⚫ 4(b) M1 any number more negative than –670 for NaCl and 4 e.g. –700 for NaCl and –750 for MgO. even more negative than that for MgO A NaCl has smaller ionic radius for M2. M2 explanation for NaCl: correct reference to I charge density unless incorrect. smaller ions / cation and anion I higher nuclear charge. M3 explanation for MgO: I atomic radius throughout. correct reference to higher charge(s) of Mg(2+) and / or O(2–) or correct reference to smaller radius / radii of cation and / or anion A comparison with NaCl or KBr for MgO. CON similar radii for M3. M4 ONE correct reference to no ecf. stronger forces / attraction between ions or stronger ionic bonds © Cambridge University Press & Assessment 2026 Page 12 of 21 Trace ID: PC-0F078E40FE Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P97a01p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/44 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 4(c)(i) number of (possible) arrangements of particles / molecules / atoms 1 I references to disorder / chaos, etc. and energy in a system I ‘arrangement of particles’ without ‘number of’. I randomness / extent. I of a substance. I ‘measure of’ for ‘number of’. I total. 4(c)(ii) decrease in number of gaseous particles OWTTE 1 A ‘gas forms solid’. A ‘more gas in reactants’ or ‘less gas in answer needs to either mention both sides of the equation or use a products’. comparative term, e.g. larger / smaller A ‘1 mol of gas in reactants and same mol e.g. ignore ‘there are no gas molecules in the products’ of solid on both sides’. but allow ‘there are fewer gas molecules in the products’ I decrease in disorder. I ‘Cl / a gas is consumed’ alone. 2 I ‘number of mols / particles decreases’. 4(c)(iii) • the reaction is less spontaneous 2 A ‘∆G more likely to be positive’ • as ∆G or (∆H – T∆S) I ∆G or ∆S becomes less exothermic becomes less negative / (more) positive I increases / decreases, etc. • due to ∆S being negative or A ΔS < 0. T∆S (becoming more) negative or I ↑ or ↓ for more / less positive. –T∆S (becoming more) positive ✓✓ ⚫ © Cambridge University Press & Assessment 2026 Page 13 of 21 Trace ID: PC-0F078E40FE Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P97a01p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/44 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/