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 shows a general increase in first ionization energy across Period 3 due to increasing nuclear charge and decreasing atomic radius. The dips at Aluminium (Al) and Sulfur (S) are due to the removal of a 3p electron (Al) and a paired 3p electron (S), respectively, which are easier to remove than the preceding elements' electrons. [xscale=0.8, yscale=0.015] % Data points for Period 3 first ionization energies (kJ/mol) % Na: 496, Mg: 738, Al: 578, Si: 787, P: 1012, S: 1000, Cl: 1251, Ar: 1521 496 738 578 787 1012 1000 1251 1521 % Axes [->] (0,0) -- (9,0) node[right] Atomic Number; [->] (0,0) -- (0,1600/15) node[above] First Ionization Energy (kJ/mol); % X-axis labels (elements) / in 1/Na, 2/Mg, 3/Al, 4/Si, 5/P, 6/S, 7/Cl, 8/Ar [below] at (,0) ; % Y-axis ticks and labels in 0,250,500,750,1000,1250,1500 (0,/15) -- (-0.1,/15) node[left] ; % Plotting the points and connecting them [blue, thick] plot[mark=o] coordinates (1,/15) (2,/15) (3,/15) (4,/15) (5,/15) (6,/15) (7,/15) (8,/15) ; (ii) Complete the table below by giving the chemical formulae of the stable oxides of the elements. | Element | Oxide Formula | | :------ | :------------ | | 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). Aluminium oxide reacts with both acids and bases. Reaction with an acid (e.g., hydrochloric acid): Al_2O_3(s) + 6HCl(aq) 2AlCl_3(aq) + 3H_2O(l) Reaction with a base (e.g., sodium hydroxide): Al_2O_3(s) + 2NaOH(aq) + 3H_2O(l) 2NaAl(OH)_4 (or Al_2O_3(s) + 2NaOH(aq) 2NaAlO_2(aq) + H_2O(l)) (v) What makes sulfur dioxide gas acidic? Sulfur dioxide (SO_2) is an acidic oxide because it reacts with water to form sulfurous acid (H_2SO_3), which then dissociates to produce hydrogen ions (H^+) in solution. SO_2(g) + H_2O(l) H_2SO_3(aq) H_2SO_3(aq) H^+(aq) + HSO_3^-(aq) 3. (b) (i) Give the outermost electronic configuration for the s-block elements (group I and group II). Group I (Alkali metals): ns^1 Group II (Alkaline earth metals): 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: Lithium has an electron configuration of 1s^2 2s^1. After the first ionization, Li^+ has an electron configuration of 1s^2, which is a stable noble gas configuration (like Helium). Removing a second electron from this very stable, fully filled inner shell requires a significantly large amount of energy. Beryllium has an electron configuration of 1s^2 2s^2. After the first ionization, Be^+ has an electron configuration of 1s^2 2s^1. The second electron is removed from the 2s orbital, which is an outer shell electron and not from a stable noble gas core. Therefore, less energy is required to remove the second electron from Be^+ compared to Li^+. Got more? Send 'em!