ASRJC 2024 H1Chem Prelim P2 Q&A

ASRJC JC2 Prelim 2024 8873/H1 [Turn over ANDERSON SERANGOON JUNIOR COLLEGE 2024 JC 2 Preliminary Paper NAME:______________________________ ( ) CLASS: 24 / ___ CHEMISTRY 8873/02 Higher 1 11 Sept 2024 2 hours Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your name, class and register number in the spaces provided at the top of this page. Write in dark blue or black pen. You may use a pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Section A Answer all questions. Section B 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. Section A 1 / 10 Section B 7/8 / 20 2 / 11 Paper 1 / 30 3 / 14 Paper 2 / 80 4 /7 Percentage Overall 5 / 7 6 /11 Grade This document consists of __ printed pages and ___blank pages.

2 ASRJC JC2 Prelim 2024 8873/H1 Section A Answer all the questions in this section in the spaces provided. 1 (a) Phosphorus, sulfur and chlorine are Period 3 elements of the Periodic Table. Table 1.1 shows some properties of the elements P to Cl. Table 1.1 P S Cl number of electrons in 3p subshell number of unpaired electrons Complete Table 1.1 to show the number of electrons in the 3p subshell and the number of unpaired electrons in an atom of P, S and Cl. [2] P [Ne]3s23p3 S [Ne]3s23p4 Cl [Ne]3s23p5 number of electrons in 3p subshell 3 4 5 number of unpaired electrons 3 2 1 (b) Fig. 1.1 shows successive ionisation energies of sulfur, S. Fig. 1.1 (i) Explain the general increase in successive ionisation energies for any atom. [2] • Nuclear charge remains unchanged for an atom 0 5000 10000 15000 20000 25000 30000 Ionisation energy / kJ mol–1 1st 2nd 3rd 4th 5th 6th 7th successive ionisation energies of S

3 ASRJC JC2 Prelim 2024 8873/H1 [Turn over • As electrons are removed from the outermost shell , there will be a decrease in shielding effect. • Electrostatic forces of attraction between the nucleus and remaining outer electrons increases. • More energy is required to remove electrons resulting in a general increase in successive ionisation energies. (ii) Complete Fig. 1.1 by plotting approximate values for the 2nd successive ionisation energy and the 6th successive ionisation energy of S. [2] (c) Describe the variation in the electrical conductivity of the elements in the third period, sodium to chlorine. Explain this variation in terms of the structures and bonding of the elements. [4] Na, Mg, Al: Giant metallic structure Presence of sea of delocalised electrons as mobile charge carriers to conduct electricity under the influence of an electric current/ field. Electrical conductivity increases across the period among the metals Number of valence electrons contributed per atom for metallic bonding increases across the period among the metals for delocalisation. More mobile electrons can act as charge carriers. Si: Giant molecular structure Strong covalent bonds exist between atoms in a giant molecular structure At higher temperatures, more electrons gain enough energy to overcome the nuclear attraction and electrical conductivity increases. Si is a 0 5000 10000 15000 20000 25000 30000 Ionisation energy / kJ mol–1 1st 2nd 3rd 4th 5th 6th 7th successive ionisation energies of S X X

4 ASRJC JC2 Prelim 2024 8873/H1 semiconductor which can only give rise to delocalised electrons at high temperatures. P4, S8, Cl2: Simple molecular structure Absence of delocalised electrons or free mobile ions as mobile charge carriers because electrons are localised in covalent bonds and not mobile to conduct electricity. They are non–conductors of electricity. [Total: 10]

5 ASRJC JC2 Prelim 2024 8873/H1 [Turn over 2 (a) When ionic compounds are dissolved in water, the ions form electrostatic attractions with water molecules. These attractions are known as ion –dipole interactions. Enthalpy change of hydration, ∆Hhyd is a measure of the strength of the ion–dipole attraction. The ionic radius of the Group 1 ions affects the strength of the attraction between the Group 1 ions and water molecules. The larger the ionic radiu s, the weaker the electrostatic forces of attraction between the ion and water molecules. Enthalpy change of hydration, ∆Hhyd is defined as the amount of heat evolved when one mole of free gaseous ions is dissolved in a large amount of water forming a solution at infinite dilution. Na+(g) Na+(aq) (i) Describe and explain how ionic radius varies down Group 1 ions. [2] Down Group 1, • The nuclear charge increases. • The number of shells increases and the valence electrons are further away from the nucleus and are more shielded. • The electrostatic forces of attraction formed between the nucleus and the valence electrons decreases, • resulting in increasing ionic radius. (ii) Hence, state the trend in the magnitude of enthalpy change of hydration, ∆Hhyd, down the Group 1 ions. [1] Magnitude of enthalpy change of hydration would decrease. (Down Group 1, ionic radius increases. Hence, the electrostatic forces of attraction between the Group 1 ion and water molecule would be weaker. Energy released on forming the ion-dipole interactions would decrease.) (iii) Suggest an equation to represent the enthalpy change of hydration, ∆Hhyd of sulfide ion, S2–. [1] S2-(g) S2-(aq) (iv) Sodium nitrate, NaNO3 is soluble in water. Draw the ‘dot-and-cross’ diagram of the nitrate ion. The ion contains nitrogen as the central atom and all atoms have 8 electrons in its outer shell. State the shape and bond angle of the nitrate ion.

6 ASRJC JC2 Prelim 2024 8873/H1 Shape: ……………………… Bond angle: ……………….. [3] N OO O . Shape: trigonal planar Bond angle: 120° (b) Table 2.1 shows the boiling point of three fluorine containing compounds. Table 2.1 formula boiling point / °C NaF 1695 HF 19.5 CH3CH2F –37.1 Explain the difference in the boiling points in terms of the structure and type of bonding in the three compounds. [4] NaF has giant ionic structure. Strong electrostatic forces of attraction exist between Na+ and F- ions as compared to the weak electrostatic forces between molecules with only partial charges. NaF has the highest boiling point since most energy is required to break the strong ionic bonds between Na+ and F- ions. CH3CH2F and HF have simple molecular structures. Weaker permanent dipole- permanent dipole (pd-pd) forces of attraction exist between CH3CH2F molecules as compared to the relatively stronger hydrogen bonds between HF molecules, resulting in least amount of energy required to overcome, hence CH3CH2F has the lowest boiling point.