ACJC 2022 Prelim P2 Answers

© ACJC2022 9729/02/Prelim/2022 [Turn over ACJC solutions for H2 Chemistry Prelim Paper 2 2022 1 (a) Table 1.1 shows the solubility of two organic molecules at 25 °C. Table 1.1 solubility in water / g dm–3 propanone miscible chloromethane 5.04 (i) Identify the type of intermolecular force present between the molecules. propanone: ………… permanent dipole -permanent dipole ……………... chloromethane:…………permanent dipole -permanent dipole ………….[1] Comments:  Some students thought that propanone has hydrogen bonding between its molecules. (ii) Give a reason for the difference in their solubilities in water Propanone is able to form hydrogen bonds with water but chloromethane is unable to. Comments:  Students need to recognise that propanone is able to form hydrogen bonds with the oxygen atom of propanone.  Students need to highlight the differences in solute-solvent interactions when discussing solubility differences.  Some students did not understand that miscible means that propanone mixes very well with water. (b) The boiling point of three ligands are shown in Table 1.2. Table 1.2 ligand formula boiling point / °C water H2O 100 ammonia NH3 –33.3 hydrazine N2H4 114 (i) Explain what is meant by the term ligand. A ligand is a neutral molecule or anion species that has a lone pair of electrons to form a dative bond with a central metal atom or ion.

© ACJC2022 9729/02/Prelim/2022 [Turn over (ii) Explain the difference in the boiling points of the three ligands. Hydrazine has the highest boiling point as it has the largest/most polarisable electron cloud size has strongest instantaneous dipole-induced dipole. Water form more extensive hydrogen bonds than ammonia and hence it has a higher boiling point. Comments:  Students need to select and compare two molecules at each time, using the appropriate concept, rather than make general statements.  Some students were confused about the electronegativity difference between nitrogen and oxygen.  Some students wrongly thought that because hydrazine being a larger molecules will be able to form more extensive hydrogen bonds. Although hydrazine does have more hydrogens atoms that can be used to form hydrogen bonds, the limiting factor is the number of lone pair of electrons. Hydrazine has the same numb er of lone pair of electrons as water (2 x 1 on each N of hydrazine and 1 x 2 on O of water). (c) Polydentate ligands are ligands which form more than one bond with the metal atom or ion. Salicylaldehyde, ethane-1,2-diamine and H2salen are examples of such ligands. H2salen can be synthesised from salicylaldehyde and ethane-1,2-diamine. When the phenolic groups of H 2salen are deprotonated, salen 2– acts as a ligand. It has a high affinity for Co2+ ions and forms a planar complex Co(salen). (i) Suggest the identity of molecule X. H2O Comments:  There is no need to write both ‘water’ and ‘H2O’ in the answer as both responses refer to the same species. Either will do.

© ACJC2022 9729/02/Prelim/2022 [Turn over (ii) Draw the structure of a salen2– ligand and circle the atoms which are used to coordinate to a Co2+ ion. salen2– [2] Comments:  In order to form the planar complex, Co(salen) needs to adopt a square planar shape about the Co 2+ centre, which indicates that there are 4 dative bonds formed. As such, four atoms should be circled.  The question asked for the salen 2– ligand only. There is no need to include Co2+ to appear in the answer. (iii) State the coordination number of the Co2+ ion in Co(salen). Four Comments:  As explained in (ii) above, to form the planar complex, Co(salen) needs to adopt a square planar shape about the Co2+ centre, which indicates that there are four dative bonds formed and hence the coordination number is four. (d) Ozonolysis is a method to oxidatively cleave alkenes using ozone, O3, to form carbonyl compounds.

© ACJC2022 9729/02/Prelim/2022 [Turn over (i) 2-methyl-3-ethylpent-2-ene Draw the structure of the organic products when 2-methyl-3-ethylpent-2-ene undergoes ozonolysis. [2] Comments:  This question was well answered by referring to reference equation. (ii) Ozonolysis of A,C6H10, gives a single compound, B,C6H10O2. B gives a yellow precipitate when treated with alkaline aqueous iodine and forms a red-brown precipitate when treated with Fehling’s solution. Draw the structures of the compounds A and B. [2] A B AA A B

© ACJC2022 9729/02/Prelim/2022 [Turn over Comments:  Students need to recognise that the alkene is part of a cyclic structure since no carbon atoms were lost.  This question proved to be demanding for many students, who did not recognise the positive iodoform test, in proposing their structure of B. [Total: 13] 2 Chromium is a transition metal that is values for its high resistance to corrosion and is added to steel to form stainless steel. While chromium can exist in various oxidation states, the most common oxidation state is the +3 state. (a) Hydrated chromium (III) chloride exists as isomers, with the general formula of CrCl3.6H2O. One such isomer is [CrCl2(H2O)4]Cl.2H2O and it appears dark green. Suggest the formula of two other isomers of hydrated chromium (III) chloride. isomer 1: ………………………………………………………………...…………….. isomer 2: …………………………………………………………………………….[2] [CrCl3(H2O)3].3H2O [CrCl(H2O)5]Cl2.H2O [Cr(H2O)6]Cl3 Comments:  To obtain the answer , students need to switch the ligands about the Cr 3+ centre.  The coordination number of chromium should not change, given that these are isomers of the given compound.

© ACJC2022 9729/02/Prelim/2022 [Turn over (b) When a sample of hydrated chromium (III) chloride is added to excess water, a green solution of CrC l3(aq) is obtained. Fig. 2.1 shows the reactions that aqueous CrCl3 can undergo. Fig. 2.1 (i) Identify precipitate A and gas B. precipitate A: ………………………………………………………………...… gas B: ………………………………………………………………………….[2] A: Cr(OH) 3 or Cr(OH) 3(H2O)3 or chromium(III) hydroxide, B: CO2 or carbon dioxide Comments:  When naming compounds, especially those involving transition metals, students need to highlight the oxidation state which the metal exists in.  Writing the chemical formula of substances is a method to identify them without ambiguity. (ii) Suggest the formula of compound C. ……………………………………………………………………………….…[1] Na3[Cr(OH)6] Comments:  Many students gave the correct anion formed without the cation.  A compound should be complete with both the cation and anion given. CrCl3 (aq) green solution Na2CO3 (aq) grey-green precipitate A + effervescence of gas B C dark green solution D violet solution excess NaOH (aq) left to stand H2O2 (aq), heat CrO42– (aq) yellow solution

© ACJC2022 9729/02/Prelim/2022 [Turn over (iii) Explain why solutions of transition metal compounds are often coloured. The d orbitals are partially filled. In the presence of ligands, the degenerate d-orbitals/d-subshell are split into two groups of different energy levels. An electron in the lower energy level orbital can absorb energy equivale