The net six-electron oxidation of aniline to nitrobenzene or azoxybenzene by cis-[Ru-IV(bpy)(2)(py)(O)](2+) (bpy is 2,2'-bipyricline; py is pyridine) occurs in a series of discrete stages. In the first, initial two-electron oxidation is followed by competition between oxidative coupling with aniline to give 1,2-diphenylhydrazine and capture by H2O to give N-phenylhydroxylamine. The kinetics are first order in aniline and first order in Ru(IV) with k(25.1 degreesC, CH3CN) = (2.05 +/- 0.18) x 10(2) M-1 s(-1) (DeltaH(double dagger) = 5.0 +/- 0.7 kcal/mol; DeltaS(double dagger) = -31 +/- 2 eu). On the basis of competition experiments, k(H2O)/k(D2O) kinetic isotope effects, and the results of an O-18 labeling study, it is concluded that the initial redox step probably involves proton-coupled two-electron transfer from aniline to cis-[Ru-IV(bpy)(2)(py)(O)](2+) (Ru-IV=O2+). The product is an intermediate nitrene (PhN) or a protonated nitrene (PhNH+) which is captured by water to give PhNHOH or aniline to give PhNHNHPh. In the following stages, PhNHOH, once formed, is rapidly oxidized by Ru-IV=O2+ to PhNO and PhNHNHPh to PhN=NPh. The rate laws for these reactions are first order in Ru-I =O2+ and first order in reductant with k(14.4 degreesC, H2O/(CH3)(2)CO) = (4.35 +/- 0.24) x 10(6) M-1 s(-1) for PhNHOH and k(25.1 degreesC, CH3CN) = (1.79 +/- 0.14) x 10(4) M-1 s(-1) for PhNHNHPh. In the final stages of the six-electron reactions, PhNO is oxidized to PhNO2 and PhN=NPh to PhN(O)=NPh. The oxidation of PhNO is first order in PhNO and in Ru-IV=O2+ with k(25.1 degreesC, CH3CN) = 6.32 +/- 0.33 M-1 s(-1) (DeltaH(double dagger) = 4.6 +/- 0.8 kcal/mol; DeltaS(double dagger) = -39 +/- 3 eu). The reaction occurs by O-atom transfer, as shown by an O-18 labeling study and by the appearance of a nitrobenzenebound intermediate at low temperature.