Zadanie 5c, Paper 2, listopad 2025, TZ3, Chemia IB SL
Wzorowane na: Chemistry SL Paper 2, November 2025, TZ3, pytanie 5(c). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
Wspólne informacje do wiązki zadań
The relative abundances of the two isotopes in a sample of chlorine are shown below.
| Isotope | Relative abundance / % |
|---|---|
| 75.50 | |
| 24.50 |
Polecenie
(i) Deduce the condensed electron configuration of chlorine.
(ii) Describe the bonding in a molecule of chlorine in terms of valence electrons.
(iii) Explain why chlorine has a much lower boiling point than bromine, although both exist as simple diatomic molecules.
(iv) Deduce the equation, and the colour change observed, for the reaction of aqueous chlorine with a dilute aqueous bromide solution.
(v) Outline why chlorine does not react with an aqueous fluoride solution.
(vi) Deduce the products of the electrolysis of molten cobalt(II) chloride, CoCl2(l).
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Odpowiedź z klucza
(i) [Ne]3s23p5 ✔
(ii) «single» shared pair «of electrons» between the two Cl atoms ✔
(iii) both Cl2 and Br2 are non-polar molecules with «only» London/dispersion forces between molecules ✔
Br2 has more electrons «greater molar mass» than Cl2, so stronger dispersion forces «between Br2 molecules, hence its higher boiling point» ✔
(iv) Cl2(aq) + 2 Br−(aq) → 2 Cl−(aq) + Br2(aq) ✔
colour change: «almost» colourless to orange/yellow ✔
(v) chlorine is a weaker oxidizing agent than fluorine / more difficult to reduce than fluorine ✔
(vi) Product at anode: chlorine / Cl2(g)
Product at cathode: cobalt / Co(s) ✔
Schemat punktowania
(i) [1]; do not accept the full, non-condensed electron configuration. (ii) [1]; do not accept just 'covalent'. (iii) M1: both molecules are non-polar with only London/dispersion forces; M2: bromine has more electrons, so stronger dispersion forces and a higher boiling point. (iv) M1: correct equation; M2: correct colour change; accept a more specific shade of orange/yellow. (v) [1]; accept a valid answer referring to oxidizing power decreasing down group 17, or to E⊖ values; do not accept just 'fluorine is more reactive' with no comparison of oxidizing/reducing power. (vi) [1] for each correct product; award [1 max] for both products correct but assigned to the wrong electrode; do not accept ions.
Komentarz. The direction of a halogen displacement reaction, and the fact that chlorine cannot displace fluoride, both follow from the same trend: oxidizing power falls down group 17, so a halogen higher in the group always displaces a halide ion below it, never the reverse.