2022 RI Prelim P2 (Ans)

1 © Raffles Institution 2022 9729/02/S/22 2022 Y6 H2 Chemistry Preliminary Exam Paper 2 – Suggested Solutions 1(a) Amount of iodine reacted = 65 22700  1 3  3 2 = 1.432  10–3 mol Mass of iodine = 1.432  10–3  126.9  2 = 0.363 g Comments: • Students need to recall that the molar volume of a gas at standard temperature and pressure (s.t.p) is 22.7 dm3 mol−1 (the molar volume at s.t.p and r.t.p can also be obtained from the Data Booklet). There is no need to use pV = nRT equation to calculate the amount of HI gas. • Using the given equations, the molar ratio of the species are as follows: amt of I2 : amt of PI3 : amt of HI 3 : 2 1 : 3 3 : 2 : 6 Therefore, amt of I2 : amt of HI 3 : 6 1 : 2 • mass of iodine = mass of I2 molecules = amt of I2  molar mass of I2 1(b)(i) The boiling points increase from HCl to HBr to HI as the size of electron cloud for polarisation increases, resulting in stronger instantaneous dipole -induced dipole (id-id) interactions, and hence requiring more energy to overcome. The boiling point decreases from HF to HCl (or HF has the highest boiling point ) due to the presence of stronger hydrogen bonding between HF molecules which requires more energy to overcome. Comments: • Boiling point is dependent on the amount of energy required to overcome the intermolecular forces of attraction . Hence, the explanation should focus on comparing the strength of attractive forces between molecules. Students should not discuss about halide ions (Cl−, Br−, I−) nor halogen atoms (Cl, Br, I). • The size and ease of polarisation of the electron cloud of HX molecule (X = Cl, Br, I) affects the strength of instantaneous dipole-induced dipole (not permanent dipole-permanent dipole) interactions between the molecules. • The strength of permanent dipole-permanent dipole interactions is not relevant in this question. Some students explained that H-Cl bond is more polar followed by H -Br and then HI. Hence, HC l molecules have the strongest pd -pd interactions followed by HBr and HI. However, the b.p of HCl < HBr < HI. • Students need to revise how hydrogen bonding arises and deduce that only HF molecule can form intermolecular hydrogen bonding.

2 © Raffles Institution 2022 9729/02/S/22 1(b)(ii) Down Group 17, the valence orbital of the halogen atom becomes increasingly diffuse and the orbital overlap between the halogen and hydrogen atoms becomes less effective / electronegativity difference between the halogen and hydrogen decreases, resulting in a decrease in bond polarity. Hence, the H–X bond strength decreases from HCl to HBr to H I and the thermal stability decreases from HCl to HBr to HI. Comments: • Thermal stability is dependent on the strength of H −X covalent bond. Hence, the explanation should focus on the effectiveness of orbital overlap between H atom and halogen atom or the polarity of the H−X covalent bond. Students should not discuss about halide ions (Cl−, Br−, I−). • Some students confused the thermal stability of Group 17 halides (H −X) with the thermal decomposition of Group 2 carbonates (M2+ CO32−) and erroneously explained about the extent of polarisation of the X− ions. Students need to take note that HX exists as simple molecules with the H−X covalent bond (not ionic bonding between H+ and X− ions). 1(c)(i) Comments: • For polyatomic anions, the extra electrons are generally gained by the more electronegative atom, e.g. oxygen atom. • Lone pairs of electrons (non -bonding electrons) in the valence shells of the central and side atoms must be shown in the ‘dot-and-cross’ diagram. • Unless otherwise stated by question, students should use only ‘dot’ or ‘cross’ to represent the valence electron from each atom and avoid using other symbols to represent the extra electron on oxygen. Between two atoms which form a bond, it should be clear to which atom (oxygen or iodine) the ‘ dot’ or ‘cross’ belongs. • Iodine is in Period 5 and can expand its octet, hence iodine should form double bonds rather than dative covalent bonds with oxygen.

3 © Raffles Institution 2022 9729/02/S/22 1(c)(ii) In IO3–, there are 3 bond pairs and 1 lone pair of electrons around the central I atom. To minimi se electronic repulsion between the bond pairs and lone pair electrons, the shape about the I atom is trigonal pyramidal. As t he lone pair-bond pair repulsion is greater than the bond pair-bond pair repulsion, the bond angle is 107o. Comments: • For questions on the shape and bond angle about a central atom, students should o state the number of bond pairs and lone pair s around the central atom in order to state/explain its shape. o compare the strength of the lone pair and/or bond pair repulsion (e.g. lone pair-bond pair vs b ond pair-bond pair repulsion) in order to state/explain the bond angle. 1(c)(iii) Method 1: Oxidation number method Oxidation number of I in I2 = 0 Oxidation number of I in IO3– = +5 Oxidation number of I in I– = –1 1 mol of I atom loses 5 mol of e– to form 1 mol of IO3–. 1 mol of I atom gains 1 mol of e– to form 1 mol of I–. Since no. of e– lost = no. of e– gained, mole ratio of IO3– : I– = 1 : 5 6OH– + 3I2 → IO3– + 5I– + 3H2O Method 2: Half-equation method Half equation from I2 to I–: I2 + 2e– → 2I– (1) Half equation from I2 to IO3–: I2 + 12OH– → 2IO3– + 6H2O + 10e– (2) Balancing equation (1) and (2): 6OH– + 3I2 → IO3– + 5I– + 3H2O Comments: • Students are reminded to give their balanced equation in the simplest mole ratio. 1(d)(i)

4 © Raffles Institution 2022 9729/02/S/22 Comments: • Curly arrows show the movement of electron pairs. • Students should compare the structures of intermediate B and the products (ethanal, butanone and HIO3) and note the bonds formed and broken in step 2 in order to suggest the mechanism. 1(d)(ii) The Ca–H bond in ethanal is stronger than that in glycol A. The 2sp2 orbital of Ca in ethanal has higher s character and is less diffuse than the 2sp3 orbital of Ca in glycol A. Hence, the 2sp2 orbital of C a in ethanal has a more effective overlap with the 1s orbital of H leading to a stronger bond. Comments: • Students should write the hybridisation properly and avoid writing sp2 or sp3. 1(d)(iii) Comments: • Students need to label their diagrams and show clearly the valence orbitals around each atom (Ca and O): o the sp2 hybrid orbitals on a trigonal plane (each sp2 orbital has a small lobe opposite to the large lobe) o the unhybridised 2p orbital which is perpendicular to the trigonal plane

5 © Raffles Institution 2022 9729/02/S/22 1(d)(iv) Comments: • Students need to apply the information given in Fig. 1.1 to work out the structure of C. Since the reaction involves the splitting of alcohols with two adjacent -OH groups into carbonyl compounds, students should focus on the two C=O groups in D and work backwards to reform the starting compound C (with two adjacent -OH groups). • Students need to read the question carefully and draw the skeletal formula. A common mistake was to include the ‘CH3’ for the methyl group. (Recall: Skeletal formula is the simplified representation of an organic formula derived from the structural formula by removing hydrogen atoms (and their associated bonds) and carbon atoms from alkyl chains, leaving just the carbon-carbon bonds in the carbon skeleton and the associated functional groups.) 1(d)(v) Nucleophilic addition HCN + OH− ⎯→ CN− + H2O Comments: • In the mechanism, students need to include the equation for the generation of