NZIC Assessment Schedule 2015 Chemistry: Demonstrate understanding of bonding, structure, properties and energy changes (91164) Assessment Criteria Evidence Statement One Expected Coverage Achievement Merit Excellence (a) H Cl C Cl Cl C Cl H In CH 2 Cl 2 there are four regions of electron density around the C atom. To minimize repulsion, these regions of electron density repel each other / are arranged as far apart as possible in a tetrahedral arrangement. Hence the bond angle is 109 o. In CCl 2 O there are three regions of electron density around the C atom. To minimize repulsion, these regions of electron density repel each other / are arranged as far apart as possible in a trigonal planar arrangement. Hence the bond angle is 120 o. O Both Lewis structures correct. Diagram of shape OR name of shape correct for one molecule. (b) (i) (ii) Br 2 O polar CO 2 non-polar O 2 non-polar Factors are The electronegativity difference between the atoms in a bond / whether the molecule has polar bonds: Cl O bond, and C = O bond, are both polar bonds as O is more electronegative than Cl, and O is more electronegative than C. However O = O bond is non polar as the atoms are identical / no electronegativity difference between the two atoms. Arrangement of polar bonds / shape of the molecule: In CO 2 is a linear molecule with the polar bonds arranged symmetrically/evenly in a linear shape around the central C atom; therefore the polar bonds/bond dipoles cancel and the molecule is non-polar. In Br 2 O, the bonds are arranged asymmetrically/unevenly in a bent shape around the central O atom; therefore the polar bonds/bond dipoles do not cancel and the molecule is polar. (O 2 is not polar as there it doesnot contain polar bonds.) H C Cl H Cl Cl O C Cl Polarity of TWO molecules correct. One of the factors correct. Explains one of the two factors. Both factors are explained, and linked to all three molecules.
One Expected Coverage Achievement Merit Excellence (c) r H o = (bonds broken) (bonds formed) Bonds broken 3 C H = 3 413 3 N H = 3 391 Total = 2412 kj mol 1 Bonds formed 3 H H = 3 436 = 1308 kj mol 1 Bonds broken and bonds formed correctly identified. Bonds broken correctly calculated. Correct calculation process to determine the bond between carbon and nitrogen with only one error. Correctly calculation to determine the bond between carbon and nitrogen linked to the correct Lewis structure. 1 C to N r H o = (bonds broken) (bonds formed) 213 = 2412 (1308 + C to N) C to N = 2412 1308 213 = 891 kj mol 1 This value matches the value for the C to N triple bond. Lewis structure is NØ N1 N2 A3 A4 M5 M6 E7 E8 No response or no relevant evidence. 1a 2a 3a 4a 1m 2m 2e with minor error 2e NZIC Level 2 Chemistry 2.4 2015 2
Two Expected Coverage Achievement Merit Excellence (a) Type of substance Type of particle Attractive forces between particles One row or column correct. Ionic ions Ionic bonds Covalent network atoms Covalent bonds between atoms in the layers Metallic Atoms / cations and electrons Metallic bonds /attraction between atom (cation) and delocalised/ valence electron (b) The forces of attractions between particles (ions / atoms in the layers / cations and electrons) are strong so a lot of energy is required to break these forces; therefore the melting points are high. Forces of attraction between particles are strong. (c) Iron atoms are held together in a 3 D lattice by metallic bonds / electrostatic attraction between atoms (cations) and delocalised / valence electrons. The attraction of the iron atoms for the delocalised electrons is not in any particular direction; therefore iron atoms can move past one another without disrupting the metallic bond, and so iron is ductile. Forces of attraction between particles are non directional. Links the structure of iron to either its ductility or conductivity Links the structure the substance to its properties (d) (i) (ii) graphite and iron For conductance there needs to be free moving charged particles. In graphite each carbon atom is bonded to three other carbon atoms in layers with the 4 th valence electron delocalised between the layers. These delocalised electrons mean that graphite can conduct electricity. In iron valence electrons of iron atoms are free to move throughout the structure. This means that iron can conduct electricity. For sodium chloride In the solid the ions are in fixed positions hence solid NaCl will not conduct electricity. When molten the ions are free to move, hence molten sodium chloride conducts electricity as ions are free to move. Structure of sodium chloride, or graphite or iron described. Graphite / Iron conducts due to delocalised electrons. Molten sodium chloride conducts due to ions that are free to move Links the structure of either sodium chloride or graphite to its conductivity. Iron to its ductility and conductivity. Graphite to its conductivity Sodium chloride to its conductivity. NZIC Level 2 Chemistry 2.4 2015 3
NØ N1 N2 A3 A4 M5 M6 E7 E8 No response or no relevant evidence. 1a 2a 3a 4a 1m 2m 2e 3e NZIC Level 2 Chemistry 2.4 2015 4
Three Expected Coverage Achievement Merit Excellence (a) (i) Exothermic heat energy is released when the temperature increases. One correct with (i) Exothermic Δ r H is negative / energy is released on combustion. reason. (b) Reaction is endothermic. When a substance changes state from a solid to a gas energy is absorbed to overcome the (weak) forces of attraction between the molecules so that they separate from each other and move apart as in the gaseous date. The solid state a number of iodine molecules close together with (weak) forces of attraction between molecules The gaseous state the iodine molecules are further apart with none/limited forces of attraction between molecules. ) Endothermic correct with reason. Weak forces between iodine molecules described either in words or in the diagram. Endothermimc reaction explained in terms of the forces of attraction in either the solid or the liquid state. Endothermimc reaction explained in terms of the forces of attraction in both the solid or the liquid state., in words and with supporting labelled diagram. (c) (i) Step ONE n(s) = m / M = 13.5 / 32 = 0.422 mol 1 mol S releases 297 kj of energy 0.422 mol S releases 0.422 x 297 = 125 kj energy Step TWO n(s) : n(so 2 ) = 0.422 mol. 2 mol SO 2 releases 99.0 kj of energy 0.422 mol S releases 0.422 99.0/2 = 20.9 kj Total energy released when 0.422 mol of S forms SO 3 = 125 + 20.9 = 146 kj Amount of S (or SO 2 ) correct Energy released for S (or SO 2 ) correct Process for calculatng total energy correct Correct calculation process to determine energy released in both steps with only one minor error. Correct calculation to determine energy in Step ONE, Step TWO and total energy released. OR Total energy released when 1 mol S forms SO 3 = 297 + 99.0/2 = 346.5 kj Total energy released when 0.422 mol S forms SO 3 = 0.422 346.5 = 146 kj NZIC Level 2 Chemistry 2.4 2015 5
NØ N1 N2 A3 A4 M5 M6 E7 E8 No response or no relevant evidence. 1a 2a 3a 4a 1m 2m 2e with minor error 2e Judgement Statement Not Achieved Achievement Achievement with Merit Achievement with Excellence Score range 0 6 7-12 13-18 19 24 NZIC Level 2 Chemistry 2.4 2015 6