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Solutions to Problems 9 (f) + Major Major H H H + (g) H:C:N:H H:C:N:H H:C::N:H + :0: :0: + H H H (Positive (Carbon Major oxygen) sextet) H H H + + + (h) H:C:C::N:O: H:C:C::N::0: H H H (Carbon sextet) Major (oxygen more electronegative than carbon) 32. Before starting, notice that the problem tells you how the atoms are attached: both compounds have two bonds. So the N is in the middle of nitromethane. We begin with σ bonds: H H-C-N O H So far, the valence shells of the carbon and the hydrogens are filled, but the nitrogen and the oxygens are short. But we have 24 electrons to work with (3 from hydrogens + 4 from C+ 5 from N + 12 from the oxygens) and we've only shown 12 in these 6 bonds. We could use the remaining 12 to add three lone pairs to each O. Let's do that, and then figure out the formal charges on the atoms: H + H-C-N H It's a "legal" Lewis structure, we've violated no rules, and we've satisfied the octets for O, but the N is in trouble with only a sextet and a 2+ charge. Can we make this better? Let's move an electron pair away from a negative atom and toward a positive one and see what we get. H H + + H-C-N H-C-N H H Now we're doing better: N has an octet, too. We could, of course, have moved an electron pair from the other oxygen instead. The result is identical to what we just got above, but the N-0 single and N=0 double bonds are switched, along with the negative charge: H H O 0: + + H C N H-C-N H H