2019 ASRJC Prelim H2 Chem P3 ANS

ASRJC JC2 PRELIMS 2019 9729/03/H2 1 ANDERSON SERANGOON JUNIOR COLLEGE H2 Chemistry 9729 2019 JC2 Prelim Exam Paper 3 Solutions 1 (a) (i) Kc = [Fe(SCN)2+] [Fe3+][SCN–] [1] units = mol–1 dm3 [1] (ii) Kc = [Fe(SCN)2+] [Fe3+][SCN–] 4.99 x 102 = [Fe(SCN)2+] [Fe3+][5.00×10–3] [Fe(SCN)2+] [Fe3+] = 2.495 [1] Let % of Fe(SCN)2+ = x %, then % of Fe3+ = 100 – x, x 100 - x = 2.495 x = 249.5 – 2.495x 3.495x = 249.5 x = 249.5 3.495 = 71.4% [1] (iii) When temperature is increased (to 500 K), backward endothermic reaction is favoured, shifting the equilibrium position to the left to decrease the temperature by absorbing the extra heat. [1] The red colour of the solution becomes less intense/fade. [1] (b) Na, Mg and Al have giant metallic lattice structure with high melting points, as a large amount of energy is required to overcome the strong electrostatic forces of attraction between cations and sea of delocalised electrons. [1] Melting point increases from Na to Al as the number of valence electrons for metallic bonding increases, and cationic radii decreases. Hence, metallic bond strength increases. [1] Si has a giant molecular structure with high melting point as very large amount of energy is required to overcome the extensive covalent bonds between atoms in the three dimensional network structure. [1] P4, S 8 and Cl2 have simple molecular structure with low melting points as small amount of energy is required to overcome the weak instantaneous dipole –induced dipole (id –id) attractions between the molecules. [1] Melting point: S8 > P4 > Cl2 o Number of electrons of S8 > P4 > Cl2 o Hence, the strength of id–id attractions between the molecules decreases across the period from S8 > P4 > Cl2 Amount of energy required to overcome the id–id attractions between the molecules decreases across the period from S8 > P4 > Cl2 [1]

ASRJC JC2 PRELIMS 2019 9729/03/H2 2 (c) (B and C could be metal oxides or chlorides due to the high melting points while D and E could be non–metal oxides or chlorides due to the low melting points.) B is SiO2 (as it has a high melting point, insoluble in water and only reacts with hot concentrated NaOH.) Reaction of B with hot NaOH: SiO2 (s) + 2OH– (aq)  SiO32– (aq) + H2O (l) C is NaCl (as it has a high melting point and dissolves in water to form a neutral solution. Due to low charge density of Na+, NaCl will not hydrolyse in water.) D has a low melting point and dissolves in water to give an acidic solution. In addition, the ratio of D to precipitated Cl– = 1: 5. D i