2024 H1 Chem JC2 Prelim P2 (Q&A) JPJC

© Jurong Pioneer Junior College [Turn over Name:____________________________________ Class:_____________ JURONG PIONEER JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION 2024 CHEMISTRY 8873/02 Higher 1 2024 Paper 2 2 hours 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. Section A (60 marks) Answer all the questions. Section B (20 marks) Answer one question. The use of an approved scientific calculator is expected, where appropriate. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use Section A 1 6 2 17 3 6 4 9 5 12 6 5 7 5 Section B 8 20 9 20 Penalty (delete accordingly) Missing/wrong units in final answer –1 / NA Total 80 This document consists of 24 printed pages and 4 blank pages.

2 © Jurong Pioneer Junior College 8873/02/J2 PRELIMINARY EXAMINATION /2024 Section A Answer all the questions in this section, in the spaces provided. 1 The oxygen family is also called the chalcogens. It consists of the elements found in Group 16 of the periodic table, oxygen, sulfur, selenium, tellurium and polonium. Chalcogen can be found in nature in both free and combined states. For Examiner’s Use (a) Complete the diagram to show the arrangement of electrons in the orbitals of a sulfur atom. 1s 2s 2p 3s 3p [1] 1s 2s 2p 3s 3p (b) State and explain the general trend in the first ionisation energy across Period 3 elements. Explain why sulfur does not follow this general trend. Across the Period, the nuclear charge increases while the shielding effect by inner shell electrons is relatively constant (since there is same number of quantum shells with electrons). Hence, the nuclear attraction on the outermost electrons increases and the 1st IE generally increases across the Period. The inter–electronic repulsion between the paired 3p electron of S makes it easier to remove one of the paired 3p electrons than the unpaired 3p electron of P which do not experience such repulsion. Hence, the 1st IE of S is lower than that of P. [2] (c) Selenium burns in air with a blue flame to form white solid SeO 2. When SeO2 reacts with water it forms H2SeO3(aq) which has a pH of less than 7. (i) Choose two oxides, each containing a different Period 3 element, which behave in a similar way to SeO2 when they are added to water. Write equation for the reaction of each oxides with water. SO2 + H2O → H2SO3 or SO3 + H2O → H2SO4 P4O6 + 6H2O → 4H3PO3 or P4O10 + 6H2O → 4H3PO4 [2] (ii) Suggest one other reagent that will react with separate samples of these oxides and with SeO2. an aqueous solution of suitable base that can react with solid acidic oxide e.g. NaOH(aq) or KOH(aq) or Ba(OH)2(aq) or Na2CO3(aq) or NaHCO3(aq) or NH3(aq) etc. [1] [Total: 6]

3 © Jurong Pioneer Junior College 8873/02/J2 PRELIMINARY EXAMINATION /2024 [Turn over 2 Nitrogen and phosphorus are elements of Group 15 in the Periodic Table. For Examiner’s Use (a) 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. Account for the difference in their physical states in terms of structure and bonding. Both N2 and P4 have simple molecular structures/ simple covalent molecules. 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 than id -id between N2. This results in higher melting point in P4, hence P4 exists as solid. [2] (b) Nitrogen exhibits a range of oxidation numbers in its compounds. (i) Complete the table below which refers to the various oxidation numbers of nitrogen in its compounds. Compound Oxidation Number of Nitrogen NO2 NO3‒ N2O NH2OH Table 2.1 [2] Compound Oxidation Number of Nitrogen NO2 +4 NO3‒ +5 N2O +1 NH2OH -1 Hydroxylamine, NH2OH is oxidised by Fe 3+(aq), which is itself reduced to Fe2+(aq). In an experiment, 0.0825 g of NH2OH required 20.00 cm3 of 0.250 mol dm-3 Fe3+ for complete reaction. (ii) How many moles of Fe3+ react with one mole of NH2OH? Amount of NH2OH reacted = 0.0825/ 33 = 2.50 x 10-3 mol Amount of Fe3+ = 20.00 x 10-3 x 0.250 = 5.00 x 10-3 mol Amount of Fe3+ reacted with one mole of NH2OH = (5.00 x 10-3) / (2.50 x 10-3) = 2 mol [1] [2] (iii) What change in oxidation number does the nitrogen in NH 2OH undergo during the reaction with Fe3+? Explain your answer.

4 © Jurong Pioneer Junior College 8873/02/J2 PRELIMINARY EXAMINATION /2024 1 Fe3+ ≡ 1e– Amount of e– gained by 2 mol of Fe3+ = 2 mol 2 mol of e– is lost by 1 mol of NH2OH. Thus, change in oxidation number in N is +2/ from -1 to +1. [1] (iv) Which formula from Table 2.1 corresponds to the nitroge n-containing product of this reaction? N2O [1] (v) Construct the half equations and hence the overall balanced equation for the reaction of NH2OH with Fe3+ (in acidic solution). [R]: Fe3+ + e– → Fe2+ [O]: 2NH2OH → N2O + H2O + 4H+ + 4e– 2NH2OH + 4Fe3+ → N2O + 4Fe2+ + H2O + 4H+ [2] (c) Nitrate, NO 3‒, and phosphate, PO 43‒, are oxoanions of nitrogen and phosphorus respectively. Phosphoric acid, H3PO4, is used in the production of Coca-Cola and other soft drinks. It serves as an acidulant, providing a tangy flavor and acting as a preservative to maintain the drink's freshness and shelf life. Draw a dot-and-cross diagram to show the bonding PO43‒. [1] (d) Phosphoryl chloride is a colourless liquid with the formula POCl₃. Table 2.2 shows the electronegativity values of the atoms in POCl₃. atom electronegativity / pauling units phosphorus 2.2 chlorine 3.0 oxygen 3.5 Table 2.2 (i) Explain the term electronegativity. Electronegativity is a measure of the tendency / ability of an atom to attract electrons in a covalent bond towards itself. Or Ability of an atom to pull the bonding pair of electrons towards itself. [1] (ii) Draw the shape of POCl₃.

5 © Jurong Pioneer Junior College 8873/02/J2 PRELIMINARY EXAMINATION /2024 [Turn over Indicate clearly the polarity of each bond it contains, and its overall net polarity. P O Cl Cl Cl net dipole moment of molecule [2] (iii) Predict all possible intermolecular forces which could exist between POCl₃ molecules. Explain how these forces arise. There are instantaneous dipole –induced dipole attractions (id–id) and permanent dipole–permanent dipole attraction (pd–pd) between POCl₃ molecules. id–id attraction arises due to the constant motion of electrons , a molecule develops an instanta neous dipole within itself when its electrons are distributed unevenly instantaneously . This instantaneous dipole then induces a temporary dipole on another molecule close to it, giving rise to id–id pd–pd attraction arises due to the electrostatic attraction between polar molecules which have permanent dipoles (or has net dipole moment within each molecule).