2021 TJC H2Chemistry P2 Ans Prelim

1 [Turn over DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN 9729 / TJC Prelim / 2021 CANDIDATE NAME CIVICS GROUP / CENTER NUMBER S INDEX NUMBER CHEMISTRY 9729/02 Paper 2 Structured Questions 24 August 2021 2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your Civics Group, centre number, index number and name in the spaces at the top of this page. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions in the spaces provided on the Question Paper. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of XX printed pages and XX blank page. For Examiner’s Use 1 2 3 4 5 Total PRELIMINARY EXAMINATIONS HIGHER 2

2 [Turn over DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN 9729 / TJC Prelim / 2021 Answer all the questions in the spaces provided. 1 (a) Excited states can be studied to gain information about the energies of orbitals that are unoccupied in an atom’s ground state. The excited state of an element, Q, is represented by the electronic configuration 1s22s22p63s23p63d44s24p1. (i) Identify the element Q. [1] Since the total number of electrons is 25, • Q is manganese. (ii) Draw the energy level diagram showing the electronic configuration of element Q in its ground state. [2] • Electronic configuration of Q in ground state - 1s22s22p63s23p63d54s2 (Either written out or shown correctly in energy level diagram) • Energy level diagram with axis and orbitals labelled, energy levels converge as distance from nucleus increase, energy diff between subshells < between shells. CLT (iii) The arrangement of electrons in the d orbitals depends on the spin states of complexes. The following diagram shows how the d orbitals are split in an octahedral environment for the ion Qn+. energy gap E d orbitals of an isolated Qn+ ion d orbitals of Qn+ ion in presence of ligands

3 [Turn over DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN 9729 / TJC Prelim / 2021 In a ‘high spin’ state, the electrons occupy all the d orbitals singly, before starting to pair up in the lower energy d orbitals. In a ‘low spin’ state, the lower energy d orbitals are filled first, by pairing up if necessary, before the higher energy d orbitals are used. Use diagrams like the one given to show the electronic configuration of a ground state Q2+ ion in low spin state. [1] Q2+: 1s22s22p63s23p63d5 • 1 mark for correct diagram (b) The second ionisation energies of seven consecutive elements A to G in the Periodic Table are shown below. (i) Write an equation for the second ionisation energy of oxygen. [1] • O+ (g) ® O2+ (g) + e- Note: State symbols must be included for IE equation. (ii) Explain the discontinuity in second ionisation energies between E and F, and between B and C. Hence deduce which element A to G is oxygen. [3] • There is a big drop in second IE from element E to F. Second electron in element F is removed from outer electronic shell. Element F is a Group 2 element (as it has two valence electrons). OR

4 [Turn over DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN 9729 / TJC Prelim / 2021 There is a big drop in second IE from element E to F. Second electron in element E is removed from inner electronic shell. Element E is a Group 1 element (as it has one valence electron). • Hence element B is oxygen (Group 16). • Second IE of C (Group 17) involves the removal of paired 2p electron which experiences inter-electronic repulsion. Hence, less energy is required to remove the paired electron from C+ ion. (c) (i) State and explain the variation in bonding in Period 3 oxides in terms of electronegativity. [2] • Across the Period, electronegativity of elements increases, electronegativity differences between the elements and oxygen decreases resulted in sharing of electrons. • The oxides become changes from ionic to covalent/increasingly covalent in character across Period 3. W, X, Y, and Z are four consecutive elements in the fourth period of the Periodic Table. The letters are not the actual symbols of the elements. W forms an oxide that reacts with both acids and bases. Z is a solid that can exist as several different allotropes. Z burns in air to form ZO2 which dissolves in water to form an acidic solution. This solution reacts with sodium hydroxide to form the salt Na2ZO3. (ii) Suggest the identities of W and Z. [1] • W is gallium, Z is selenium. (iii) Write equations for the reactions of oxide of W with sodium hydroxide and hydrochloric acid respectively. [2] • W2O3 + 2NaOH + 3H2O ® 2NaW(OH)4 • W2O3 + 6HCl ® 2WCl3 + 3H2O (iv) Write an equation for the formation of an acidic solution when ZO2 dissolves in water. [1] • ZO2 + H2O ® H2ZO3 [Total: 14]

5 [Turn over DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN 9729 / TJC Prelim / 2021 2 (a) Chlorine reacts with iodine to form a compound T, ICl7. When dissolved in an excess of aqueous potassium iodide, T liberates iodine, I2, which is the only iodine-containing product in the reaction. (i) State the oxidation number of iodine in ICl7. [1] • +7 (ii) Write the equation for the reaction between T and potassium iodide. [1] • ICl7 + 7KI 4I2 + 7KCl +7 7(-1) 0 (iii) Calculate the amount of iodine liberated when 1.00 g of T reacts with an excess of aqueous potassium iodide. [1] Amount of ICl7 = 1/(126.9 + 7 x 35.5) = 2.66 x 10–3 mol Amount of iodine = 4 x 2.66 x 10–3 • = 1.07 x 10–2 mol (allow ecf from (a)(ii)) (iv) Sodium thiosulfate is a common reagent used for the reaction with iodine. Write a balanced equation for the reaction between sodium thiosulfate and iodine and calculate the volume of 1.00 mol dm–3 sodium thiosulfate, in cm3, required to react with all the iodine liberated in (a)(iii). [2] • I2 + 2S2O32– 2I– + S4O62– Volume of sodium thiosulfate required = (1.07 x 10–2 x 2)/1 x 1000 = • 21.3 cm3 (allow ecf from a(iii)) (b) Chlorine is produced together with carbon monoxide when phosgene, COCl2, undergoes dissociation according to the equation below: COCl2(g) ⇌ Cl2(g) + CO(g) The above reaction takes plac