VJC 2021 Prelim H2P3 Ans [Final]

1 Ó VJC 2021 9729/03/PRELIM/21 [Turn over Victoria Junior College 2021 H2 Chemistry Prelim Exam 9729/3 Suggested Answers Section A Answer all the questions in this section. 1 (a) Elements in the same group tend to form products with similar physical properties. However, the oxides of carbon and silicon have very different physical structures. While CO2 exists as a gas at room temperature, SiO2 is a solid. (i) Explain, in terms of structure and bonding, the difference in the physical states of CO2 and SiO2. SiO2 has a giant molecular structure where each Si atom is covalently bonded to 4 other O atoms. CO2 has simple molecular structure with weak instantaneous dipole-induced dipole interactions (id-id) between the CO2 molecules. Much more energy is required to break the strong covalent bonds in SiO2 as compared to the weak id-id for CO2. Hence, SiO2 is a solid while CO2 is a gas at room temperature. [2] (ii) Describe the bonding in a single CO2 molecule. Include a fully labelled diagram showing how the different types of bonds are formed in terms of orbital overlap. C–O σ bond is formed via head-on overlap between sp hybridized orbital of C and sp2 hybridised orbital of O (accept p orbital of O). C–O π bond is formed via side-ways overlap between unhybridized p orbital of C and unhybridised p orbital of O. [2] (iii) By considering the extent of overlap of the orbitals, suggest a reason why SiO2 does not form bonds similar to that of CO2. Si atom has a larger atomic radius than C. This leads to less effective overlap of the p orbitals between Si and O. Thus, a π bond formed between S and O would be much weaker and less likely to be formed. [1] σ bond π bond sp2 orbital of O sp orbital of C p orbital of C p orbital of O

2 Ó VJC 2021 9729/03/PRELIM/21 [Turn over (b) The extraction of iron from its ores typically involves the use of a Blast Furnace. One of the most common impurities in iron ore is silicon dioxide. To remove silicon dioxide, limestone is added. The calcium carbonate in limestone decomposes due to the high temperatures in the Blast Furnace to form calcium oxide. The calcium oxide then reacts with silicon dioxide to form calcium silicate, Ca2SiO4, which runs to the bottom of the Blast Furnace to be removed easily. (i) State the type of reaction between silicon dioxide and calcium oxide. acid-base [OR neutralization] [1] (ii) Define the term lattice energy. Lattice energy is energy released when one mole of solid ionic solid is formed from its isolated gaseous ions. [1] (iii) Using the information given in Table 1.1 and relevant data from the Data Booklet, draw an energy level diagram and calculate a value for the lattice energy of calcium silicate, Ca2SiO4(s). Table 1.1 enthalpy change value/kJ mol–1 standard enthalpy change of atomisation of Ca(s) +178 standard enthalpy change of atomisation of Si(s) +456 O(g) + 2e– ® O2–(g) +650 Si4+(g) + 4O2–(g) ® SiO44–(g) –9304 standard enthalpy change of formation of Ca2SiO4(s) –994

3 Ó VJC 2021 9729/03/PRELIM/21 [Turn over L.E. = 994 + 2(178) + 456 + 2(496) + 2(590 + 1150) + (789 + 1580 + 3230 + 4360) + 4(650) – 9304 LE = –9530 kJ mol–1 [4] (iv) The reaction between calcium oxide and silicon dioxide also forms another form of calcium silicate, CaSiO3. Deduce whether the magnitude of the lattice energy of CaSiO3 is expected to be larger or smaller than that of Ca2SiO4. Explain your answer. L.E. ∝𝐪!∙𝐪–𝐫!$𝐫– The cationic charge and cationic radius are the same for both compounds. SiO32– has a smaller anionic charge and smaller ionic radius than SiO44–. Charge factor dominates and hence, CaSiO3 will have a lattice energy with smaller magnitude. [2] 2Ca(s) + Si(s) + 2O2(g) 2Ca(g) + Si(g) + 2O2(g) 2Ca2+(g) + Si4+(g) + 4O(g) + 8e 2Ca(g) + Si(g) + 4O(g) 2Ca2+(g) + Si4+(g) + 4O2–(g) Ca2SiO4(s) 2(+178) + 456 2(+496) 0 energy, kJ mol–1 2(+590+1150) + (+789 +1580+3230+4360) 4(+650) –994 –9304 2Ca2+(g) + SiO44–(g) L.E.

4 Ó VJC 2021 9729/03/PRELIM/21 [Turn over (c) Compound A, C9H9NO2, is insoluble in both aqueous acids and alkalis and does not react with sodium metal. When heated with aqueous hydrochloric acid, followed by careful addition of aqueous sodium hydroxide, compound B, C8H11NO is formed. Gentle warming of A with acidified KMnO4 gives the salt of compound C, C8H9NO. Further strong heating of the mixture forms the salt of compound D, C7H7NO2. Each mole of compounds B, C and D decolourises two moles of aqueous bromine. Compound C gives a pale yellow precipitate on warming with alkaline aqueous iodine. Suggest structures for A, B, C and D and explain the reactions described. Hints Deductions 1 Compound A has C:H = 1:1 A contains a benzene ring. 2 Compound A, C9H9NO2, is insoluble in both aqueous acids and alkalis and does not react with sodium metal. A is a neutral compound. A does not contain –NH2 or –COOH. A contains amide / ester. A does not contain alcohol, phenol or carboxylic acid. 3 When heated with aqueous hydrochloric acid, followed by careful addition of NaOH, compound B, C8H11NO is formed. A undergoes acid hydrolysis of amide (and ester) followed by neutralisation. B contains an amine (and alcohol). Loss of carbon suggests the carboxyl carbon was removed upon hydrolysis. 4 Gentle warming of A with acidified KMnO4 gives C, C8H9NO. A undergoes hydrolysis to give B which undergoes oxidation of 2º alcohol to give ketone C 5 Further strong heating of the mixture forms D, C7H7NO2. Loss of carbon suggests C undergoes side–chain oxidation to give benzoic acid in D. 6 Each mole of compounds B, C and D decolourises two moles of aqueous bromine. B, C and D undergoes electrophilic substitution. They contain phenylamine. The compounds are either 1,2 or 1,4 substituted. 7 Compound C gives a pale yellow precipitate on warming with alkaline aqueous iodine. C undergoes oxidative cleavage to give CHI3 yellow ppt. C contains –COCH3 structural unit. A: B: C: D: [8] [Total: 21]

5 Ó VJC 2021 9729/03/PRELIM/21 [Turn over 2 (a) (i) The reaction between an aqueous solution of permanganate(VII) ions, MnO4–, and ethanedioate ions, C2O42–, can be represented by the following equation: 5C2O42–(aq) + 2MnO4–(aq) + 16H+(aq) → 10CO2(g) + 2Mn2+(aq) + 8H2O(l) The presence of Mn2+, which are produced in the reaction between MnO4– and C2O42–, is thought to catalyse this reaction. An experiment was performed to measure the volume of CO2 gas produced at regular time intervals until the reaction goes to completion. The graph of volume of CO2 gas produced against time has three distinct regions. Sketch on Figure 2.1, the graph of volume of CO2 gas produced against time you would expect to obtain. Figure 2.1

6 Ó VJC 2021 9729/03/PRELIM/21 [Turn over Figure 2.1 [1] (ii) Explain why there are three distinct regions of the graph you have drawn in (a)(i). Initially, the rate of reaction will be slow due to the electrostatic repulsion between the negatively charged MnO4– and C2O42–, which results in high activation energy to overcome. Thus, the gradient will be gentle. As the reaction pr