2022 JPJC prelim P2 Answer

© Jurong Pioneer Junior College [Turn Over NAME CLASS 21S JURONG PIONEER JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION 2022 CHEMISTRY 9729/02 Higher 2 Paper 2 Structured Questions 14 September 2022 2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your name, class and exam index number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a HB pencil for any diagrams, 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. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 10 2 10 3 17 4 10 5 11 6 12 7 5 Penalty (delete accordingly) Bond linkages –1 / NA Significant figures & units –1 / NA Total 75 This document consists of 20 printed pages.

2 © Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2022 Answer all the questions in this section. 1 (a) Figure 1.1 shows the third ionisation energies of eight consecutive elements A to H, in the Periodic Table. [Note that letters A to H are not the atomic symbols of the elements concerned.] Figure 1.1 For Examiner’s Use (i) Write an equation, including state symbols, to represent the third ionisation energy of element A. [1] A2+(g) ⟶ A3+(g) + e- (ii) From Figure 1.1, suggest the identity of B. Explain how you arrived at your answer. [2] The large decrease in 3 rd IE from F to G implies that G 2+ has 1 more quantum shell of electrons than F2+ / 3rd electron to be removed from F is from the inner quantum shell. The valence shell configuration of G2+ is ns1 and hence, the valence shell configuration of G is ns2 np1 (i.e. Al). OR This also implies that the valence shell configuration of B2+ is (n-1)s2 np2 and hence, the valence shell configuration of B is ns2 np4 . OR F is in group 2 / G is in group 13 Hence, B is oxygen. Cannot accept sulfur. (iii) Explain why the third ionisation energy of element D is slightly lower than that of element C. [1] C2+: 1s22s22p3 D2+: 1s22s22p4 The inter −electronic repulsion / repulsion between the paired 2p electrons in D 2+ makes it easier to remove a paired electron than an unpaired 2p electron of C2+ which do not experience such repulsion. Hence, the 3rd IE of D is lower than that of element C. Third Ionisation energy/ kJ mol−1

3 © Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2022 [Turn Over (b) Nitrogen and phosphorus are elements of Group 15 in the Periodic Table. Nitrogen exists naturally as gaseous diatomic N N molecules whereas phosphorus is a solid and exists as P 4 molecules comprising of P -P single bonds. (i) Account for the difference in their physical states in terms of structure and bonding. [2] Both N2 and P4 have simple molecular structures. As P 4 has larger number of electrons/ bigger electron cloud to be polarised, more energy is required to overcome the stronger instantaneous dipole-induced dipole interactions/attractions between the P4 molecules. This results in higher melting point in P4, hence P4 exists as solid. (ii) Suggest why phosphorus does not occur naturally as PP molecules. [1] Phosphorus is a relatively big atom with diffused orbitals, side-on overlap of its p orbitals to form π bonds is much less effective than head -on overlap to form sigma bond. (iii) Nitrate, NO 3‒, and phosphate, PO 43‒, are oxoanions of nitrogen and phosphorus respectively. Draw a dot-and-cross diagram to show the bonding PO43‒, deducing the shape and the bond angle around the phosphorous atom. Hence explain why it is not possible for nitrogen to form an oxoanion with formula of NO43‒. [3] 4 bond pairs around central atom P Shape of PO43‒ is tetrahedral, angle O—P—O: 109.5o To form NO 43‒, N must be able to accommodate 10 electrons in its valence shell. Since N is in Period 2, it has no energetically accessible/low lying d orbital to expand its octet. [Total: 10]

4 © Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2022 2 Iron oxides are chemical compounds composed of iron and oxygen. Most iron ores are oxides, making them important precursors to iron metal and its many alloys. (a) Iron ( II) compounds are generally only stable in neutral, non -oxidising conditions. It is difficult to determine the lattice energy of FeO experimentally. (i) Given the following data in Table 2.1 and data from the Data Booklet, use the energy diagram below to calculate the lattice energy of FeO(s) in kJ mol-1. standard enthalpy change of atomisation of Fe(s) +416 kJ mol–1 standard enthalpy change of formation of FeO(s) –272 kJ mol–1 Sum of 1st and 2nd Electron Affinity of oxygen +157 kJ mol–1 [3] Table 2.1 By Hess Law, ‒272 = +416 + ½ (496) + 762 + 1560 +157 + LE (FeO) LE (FeO) = ‒3415 (or -3420) kJ mol-1 FeO(s) 0 1st + 2nd I.E. (Fe) 1/2 E(O=O) 1st + 2nd E.A. (O) LE (FeO) Enthalpy ‒272 kJ mol-1 Fe (s) + ½O2 (g) Fe (g) + ½O2 (g) +416 kJ mol-1 Fe2+ (g) + O2‒ (g) Fe (g) + O (g) Fe2+(g) + 2e- + O(g)

5 © Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2022 [Turn Over (ii) Most naturally occurring samples of iron( II) oxides are found as the mineral WÜstite. WÜstite has the formula Fe20Ox. It contains both Fe2+ and Fe3+ ions. 90% of the iron is present as Fe2+ and the remaining as Fe3+. Deduce the value of x. [1] Balancing of charges: ( ) ( ) + + + − =20 0.9 2 0.1 3 2 0 x Solving x = 21 (iii) State and explain how the lattice energies of FeO compares with the lattice energies of Fe2O3. [2] Fe3+ ion has a higher (ionic) charge and smaller radius/size than Fe2+ ion. Since +− +−  + LE qq rr , hence, the magnitude of LE for Fe2O3 is larger / LE for Fe2O3 is more exothermic than that in FeO. (b) Another common iron oxide as hematite, Fe2O3 is the main source of iron for the steel industry. Fe2O3 will readily react with acids to form soluble salts such as the following reaction. Fe2O3(s) + 6HI(aq)→ 2FeI2(aq) + I2(aq) + 3H2O(l) (i) Define standard enthalpy change of solution. [1] It is the heat change when 1 mole of a substance is completely dissolved in water under the standard conditions of 298 K and 1 bar so that there is no further heat change upon adding more water. (ii) Use the data in Table 2.2 to calculate the enthalpy change of solution of iron (II) iodide, FeI2. ∆Hlatt (FeI2(s)) ‒ 2440 kJ mol‒1 ∆Hhyd (Fe2+ (g)) ‒ 1950 kJ mol‒1 ∆Hhyd (I‒ (g)) ‒ 308 kJ mol‒1 [1] Table 2.2 Hsoln = −(−2440) + (−1950) + (2(−308)) = − 126 kJ mol-1 (iii) A yellow precipitate of PbI2 forms when 25 cm 3 of x mol dm-3 Pb2+ ions are added to 10 cm3 of 0.100 moldm-3 FeI2 (aq). Given that the solubility product, Ksp, of PbI2 = 9.8 x 10 -9 mol3 dm-9, find the minimium value for x, concentration of Pb2+. [2] For ppt to occur, Ionic product > Ksp [Pb2+] [I‒]2 [Pb2+] [I‒]2 > 9.8 x 10-9

6 © Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2022 25 1000 35 1000 x     2 102 0.1001000 35 1000      > 9.8 x 10-