The reaction of phenyl azide with (PNP)Ni, where PNP = ((Bu2PCH2SiMe2)-Bu-t)(2)N-, promptly evolves N-2 and forms a P=N bond in the product (PNP=NPh)Ni-I. A similar reaction with (PNP)FeCl proceeds to form a P=N bond but without N-2 evolution, to furnish (PNP=N-N=NPh)FeCl. An analogous reaction with (PNP)RuCl occurs with a more dramatic redox change at the metal (and N-2 evolution), to give the salt composed of (PNP)Ru(NPh)(+) and (PNP)RuCl3-, together with equimolar (PNP)Ru(NPh). The contrast among these results is used to deduce what conditions favor N2 loss and oxidative incorporation of the NPh fragment from PhN3 into a metal complex.