1 (276)

Prévia do material em texto

9 Molecular Geometry Solutions to Exercises (c) 14 valence pairs H H 4 electron domains around N, sp³ hybridization N₂H₂, 12 valence e⁻, 6e⁻ pairs 3 electron domains around N, hybridization N₂, 10 valence 5 e- pairs :N=N: 2 electron domains around N, sp hybridization (d) In the three types of N-N bonds, each N atom has a nonbonding or lone pair of electrons. The lone pair to bond pair repulsions are minimized going from 109° to 120° to 180° bond angles, making the π bonds stronger relative to the σ bond. In the three types of C-C bonds, no lone-pair to bond-pair repulsions exist, and the σ and π bonds have more similar energies. H 9.114 H H (g) 6C(g) + 6H(g) H H H AH = 6D(C-H) + 3D(C-C) + 3D(C=C) - 0 = 6(413 kJ) + 3(348 kJ) + = 5364 kJ (The products are isolated atoms; there is no bond making.) According to Hess' law: = C(g) + H(g) - C₆H₆(g) = 6(718.4 + 6(217.94 - (82.9 = 5535 kJ The difference in the two results, 171 kJ/mol C₆H₆ is due to the resonance stabilization in benzene. That is, because the π electrons are delocalized, the molecule has a lower overall energy than that predicted for the presence of 3 localized C-C and C=C bonds. Thus, the amount of energy actually required to decompose 1 mole of C₆H₆(g), represented by the Hess' law calculation, is greater than the sum of the localized bond enthalpies (not taking resonance into account) from the first calculation above. 9.115 (a) (b) Ignoring the donut of the orbital 270