Wzorowane na: Chemistry HL Paper 2, May 2025, TZ2, pytanie 1(a). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
Informacja do zadań 1a-1d.
Hydrogen cyanide,
At room temperature hydrogen cyanide is a liquid that evaporates readily, whereas ethyne, , 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) Deduce the hybridization of the nitrogen atom in and the number of sigma and pi bonds that this atom forms.
(iii) State the molecular geometry of and account for it using the VSEPR model, without referring to hybridization.
(iv) 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.
(v) Explain, in terms of intermolecular forces, why the boiling point of ethyne is so much lower than that of hydrogen cyanide.
(i) drawn with one shared pair between H and C, three shared pairs between C and N and one lone pair on N «10 valence electrons in total; dots, crosses or lines accepted» ✔
(ii) Hybridization of N: sp ✔
Sigma bonds: 1 AND pi bonds: 2 «the lone pair occupies the second sp orbital» ✔
(iii) 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° ✔
(iv) Partial negative charge: N «electronegativity 3.0» AND partial positive charge: H «electronegativity 2.2; accept C» ✔
(v) 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 ✔
(i) [1] complete Lewis formula including the lone pair on N. (ii) [2] M1 sp; M2 one sigma and two pi bonds, both needed. (iii) [2] M1 linear; M2 VSEPR reasoning: two electron domains or two bonding pairs and no lone pairs on C, maximum separation. (iv) [1] both partial charges correct; accept C for the partial positive charge. (v) [2] M1 identifies the intermolecular forces in each substance; M2 compares their strength. Do not accept 'van der Waals forces' for M1.
In (ii) the terminal nitrogen atom uses one sp orbital for the sigma bond and the other for its lone pair, so it forms only one sigma bond even though it is sp hybridized.
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