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8 Chemical Bonding Solutions to Exercises (b) SrCl₂(s) = Cl(g) + 2E(CI) - SrCl₂ SrCl₂(s) = 164.4 kJ + 549 kJ + 1064 + 2(121.7) kJ + kJ 2127 kJ = -804 kJ 8.103 The pathway to the formation of can be written: 2K(s) 2K(g) K(g) 2K(g) (K) O(g) O(g) O(g) + 1 1 E₁ (O) (g) + 1 -> K₂O(s) K₂O(s) K₂O(s) K₂O(s) = O(g) + latt K₂O(s) (O) = K₂O(s) + latt K₂O(s) E₂(O) = -363.2kJ + 2238 kJ 2(89.99) kJ 2(419) kJ - 247.5 kJ - (-141) kJ = +750 kJ 8.104 To calculate empirical formulas, assume 100 g of sample. 76.0 Ru (a) = 0.752 mol Ru; 0.752/0.752 = 1 Ru 101.07g/mol 24.0 g = 1.50 mol O; 1.50/0.762 = 2 The empirical formula of compound 1 is RuO₂. (b) 61.2 Ru = 0.6055 mol Ru; 0.6055/0.6055 = 1 Ru 101.07 / mol 38.8 g = 2.425 mol O; 2.425/0.6055 = 40 15.9994 g / mol The empirical formula of compound 2 is (c) Ionic compounds have very high melting points, while the melting points of molecular compounds are lower and variable. Clearly the black powder, m.p. > 1200°C, is ionic and the yellow substance, m.p. = 25°C, is molecular. Substances with metals in high oxidation states are often molecular. contains Ru(VIII), while RuO₂ contains Ru(IV), so RuO₄ is more likely to be molecular. The yellow compound is ruthenium tetroxide. The black compound is ruthenium(IV) oxide. 8.105 (a) Even though has the greater (more negative) electron affinity, F has a much larger ionization energy, so the electronegativity of F is greater. F: k(IE-EA) = k(1681 (-328)) = k(2009) Cl: k(IE-EA) = k(1251 (-349)) = k(1600) 224