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8 Chemical Bonding Solutions to Exercises 8.93 (a) 12 + 3 + 15 = 30 valence e-, 15 pairs. H H H H C C C C :N N: :N N: C C H H Structures with H bound to N and nonbonded electron pairs on C can be drawn, but the structures above minimize formal charges on the atoms. (b) The resonance structures indicate that triazine will have six equal C-N bond lengths, intermediate between C-N single and C-N double bond lengths. (See Solutions 8.57 and 8.58.) From Table 8.5, an average C-N length is 1.43 Å, a C=N length is 1.38 Å. The average of these two lengths is 1.405 Å. The C-N bond length in triazine should be in the range 1.40-1.41 Å. 8.94 (a) 24 + 4 + 14 = 42 valence 21 pairs. (b) H H H H C C H-C C-H H H C=C Cl (c) In benzene, the six C atoms are equivalent. In ortho-dichlorobenzene, the two C atoms bound to Cl are not equivalent to the four C atoms bound to H. In the two resonance structures above, one has a double bond between the C atoms bound to and the other has a single bond in this position. The two ortho- dichlorobenzene resonance structures are not equivalent like the resonance structures of benzene. 8.95 An experimentally determined molecular structure will reveal bond lengths and angles of the B-A=B molecule. If resonance structures are important, the two B-A bond lengths will be identical, or nearly so. If the molecule features one single and one double bond, the lengths will be significantly different. (While bond angles often reveal bonding details, in this case bond lengths are telling.) 8.96 + = 2(413) + 1/2 (495) 348 2(463) = -200 kJ The fundamental difference in the two reactions is the formation of 1 mol of H-H bonds versus the formation of 2 mol of O-H bonds. The latter is much more exothermic, so the reaction involving oxygen is more exothermic. 220