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8 Chemical Bonding Solutions to Exercises 8.6 (a) 18 valence e-, pairs 18 valence 9 e- pairs H-O-N=O (b) The formal charge on N is zero, in both species. (c) is expected to exhibit resonance; the double bond can be drawn to either oxygen atom. An alternate resonance structure for can be drawn, but it has nonzero formal charges on the oxygen atoms. This structure is less likely than the one shown above. (d) Assuming that the structure shown above is the main contributor to the structure of the N=O bond length in HNO₂ will be shorter than the N-O lengths in Because there are two equivalent resonance structures for the N-O lengths are approximately an average of N-O single and double bond lengths. These are longer than the full N=O double bond in 8.7 Analyze/Plan. Since there are no unshared pairs in the molecule, we use single bonds to H to complete the octet of each C atom. For the same pair of bonded atoms, the greater the bond order, the shorter and stronger the bond. Solve. (a) Moving from left to right along the molecule, the first C needs two H atoms, the second needs one, the third needs none, and the fourth needs one. The complete molecule is: H H 1 2 3 (b) In order of increasing bond length: 3