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9 Molecular Geometry Solutions to Exercises (b) A bond order of one corresponds to a single bond, which is a σ bond; there is free rotation about a σ bond. Absorption of a photon of appropriate wavelength breaks the π bond in ethylene but leaves the σ bond intact; it reduces the bond order from two to one. There is facile rotation about the remaining carbon- carbon σ bond. Integrative Exercises 9.109 (a) Assume 100 g of compound 2.1g H 1molH = 2.1 mol H; 2.1/2.1 = 1 29.8gN x 14.01gN 2.13 mol N; 2.13/2.1 ≈ 1 1molO mol O; 4.26/2.1 ≈ 2 The empirical formula is formula weight = 47. Since the approximate molar mass is 50, the molecular formula is (b) Assume N is central, since it is unusual for to be central, and part (d) indicates as much. HNO 18 valence e- -1 0 +1 The second resonance form is a minor contributor due to unfavorable formal charges. (c) The electron domain geometry around N is trigonal planar with an O-N-O angle of approximately 120°. If the resonance structure on the right makes a significant contribution to the molecular structure, all four atoms would lie in a plane. If only the left structure contributes, the H could rotate in and out of the molecular plane. The relative contributions of the two resonance structures could be determined by measuring the O-N-O and N-O-H bond angles. (d) 3 VSEPR e- domains around N, hybridization (e) 1 π for both structures (or for H bound to N). 9.110 (a) (g) + (g) (g) (b) 40 20 pairs There must be a double bond drawn between o and S in order for their formal charges to be zero. (c) + = D(O=O) - 2D(S=O) = 495 2(523) = -551 kJ, exothermic 267