Zadanie 1a, Paper 2, maj 2025, TZ2, Chemia IB SL
Wzorowane na: Chemistry SL Paper 2, May 2025, TZ2, pytanie 1(a). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
Wspólne informacje do wiązki zadań
Hydrogen cyanide, HCN, is an extremely poisonous compound that is nevertheless manufactured on a large scale, because it is a starting material for polymers such as nylon and poly(methyl methacrylate).
Polecenie
At room temperature hydrogen cyanide is a liquid that evaporates readily, whereas ethyne, C2H2, which has almost the same molar mass, is a gas that boils at −84 °C.
(i) Draw the Lewis formula of the hydrogen cyanide molecule, showing all valence electrons.
(ii) State and explain the molecular geometry of HCN, using the VSEPR model.
(iii) HCN has a permanent dipole. Use section 9 of the data booklet to deduce which atom carries the partial negative charge and which atom carries the partial positive charge.
(iv) Explain, in terms of intermolecular forces, why the boiling point of ethyne is so much lower than that of hydrogen cyanide.
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Odpowiedź z klucza
(i) H−C≡N drawn with one shared pair between H and C, three shared pairs between C and N and one lone pair on N «ten valence electrons in total; dots, crosses or lines accepted» ✔
(ii) Molecular geometry: linear ✔
Explanation: the central carbon atom has two electron domains «one single bond and one triple bond» and no lone pairs, and the two domains repel each other to the greatest possible separation, 180° ✔
(iii) Partial negative charge: N «electronegativity 3.0» AND partial positive charge: H «electronegativity 2.2; accept C» ✔
(iv) ethyne is non-polar, so its molecules are held together only by London (dispersion) forces, whereas HCN is polar and has dipole–dipole attractions «as well as London forces» ✔
dipole–dipole attractions are stronger than London forces between molecules of similar molar mass, so more energy is needed to separate HCN molecules ✔
Schemat punktowania
(i) [1] complete Lewis formula including the lone pair on N. (ii) [2] M1 linear; M2 VSEPR reasoning: two electron domains or two bonding pairs and no lone pairs on C, maximum separation. (iii) [1] both partial charges correct; accept C for the partial positive charge. (iv) [2] M1 identifies the intermolecular forces in each substance; M2 compares their strength. Do not accept 'van der Waals forces' for M1.