This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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You're on a roll — SECTION B: INORGANIC CHEMISTRY 3. (a) This question concerns the elements across period 3 of the periodic table (Na - Ar). (i) Sketch a graph for the variation of first ionization energy against atomic number for the elements. The graph generally shows an increasing trend in first ionization energy across Period 3, with notable dips at Aluminium (Al) and Sulfur (S). X-axis: Atomic Number (11 to 18) Y-axis: First Ionization Energy (kJ/mol) Points to plot (approximate relative values): Na (11): Low Mg (12): Higher than Na Al (13): Lower than Mg (due to 3p orbital shielding) Si (14): Higher than Al P (15): Higher than Si S (16): Lower than P (due to electron-electron repulsion in half-filled 3p orbital) Cl (17): Higher than S Ar (18): Highest ` First Ionization Energy (kJ/mol) ^ | Ar | Cl | P | Si | S | Mg | Al | Na +-------------------------------------> Atomic Number 11 12 13 14 15 16 17 18 Na Mg Al Si P S Cl Ar ` (ii) Complete the table below by giving the chemical formulae of the stable oxides of the elements. | Element | Oxide | | :------ | :---- | | Na | Na_2O | | Mg | MgO | | P | P_4O_10 | | Cl | Cl_2O_7 | (iii) Write a balanced equation for the reaction of silicon tetrachloride with water. SiCl_4(l) + 2H_2O(l) SiO_2(s) + 4HCl(aq) (iv) Write two equations to show the amphoteric nature of aluminium oxide (Al_2O_3). 1. Reaction with an acid (e.g., hydrochloric acid): Al_2O_3(s) + 6HCl(aq) 2AlCl_3(aq) + 3H_2O(l) 2. Reaction with a base (e.g., sodium hydroxide): Al_2O_3(s) + 2NaOH(aq) + 3H_2O(l) 2NaAl(OH)_4 (v) What makes sulphur dioxide gas acidic? Sulphur dioxide (SO_2) is an acidic gas because it reacts with water to form sulfurous acid (H_2SO_3), which can then donate protons. SO_2(g) + H_2O(l) H_2SO_3(aq) 3. (b) (i) Give the outermost electronic configuration for the s-block elements (group I and group II). Group I: ns^1 Group II: ns^2 (ii) For the elements Lithium and Beryllium, write equations representing: A. First ionization energy for Lithium Li(g) Li^+(g) + e^- B. Second ionization energy for Beryllium Be^+(g) Be^2+(g) + e^- (iii) Which of these two elements in b(ii) will have a higher second ionization energy? Explain. Lithium will have a higher second ionization energy. Explanation: After losing its first electron, Lithium (Li) forms Li^+, which has a stable electron configuration similar to Helium (1s^2). Removing a second electron from Li^+ requires breaking into this very stable, full inner electron shell, which demands a significantly large amount of energy. Beryllium (Be), after losing its first electron, forms Be^+, which still has one electron in its 2s orbital (1s^2 2s^1). Removing this second electron from Be^+ is easier than removing an electron from the stable 1s^2 core of Li^+. Drop the next question.