The feasibility of oxidative addition of the P-H bond of PHPh2 to a series of rhodium complexes to give mononuclear hydrido-phosphanido complexes has been analyzed. Three main scenarios have been found depending on the nature of the L ligand added to [Rh(Tp)(C2 H4 )(PHPh2 )] (Tp= hydridotris(pyrazolyl)borate): i) clean and quantitative reactions to terminal hydrido-phosphanido complexes [RhTp(H)(PPh2 )(L)] (L=PMe3 , PMe2 Ph and PHPh2 ), ii) equilibria between RhI and RhIII species: [RhTp(H)(PPh2 )(L)]⇄[RhTp(PHPh2 )(L)] (L=PMePh2 , PPh3 ) and iii) a simple ethylene replacement to give the rhodium(I) complexes [Rh(κ2 -Tp)(L)(PHPh2 )] (L=NHCs-type ligands). The position of the P-H oxidative addition-reductive elimination equilibrium is mainly determined by sterics influencing the entropy contribution of the reaction. When ethylene was used as a ligand, the unique rhodaphosphacyclobutane complex [Rh(Tp)(η1 -Et)(κC,P -CH2 CH2 PPh2 )] was obtained. DFT calculations revealed that the reaction proceeds through the rate limiting oxidative addition of the P-H bond, followed by a low-barrier sequence of reaction steps involving ethylene insertion into the Rh-H and Rh-P bonds. In addition, oxidative addition of the P-H bond in OPHPh2 to [Rh(Tp)(C2 H4 )(PHPh2 )] gave the related hydride complex [RhTp(H)(PHPh2 )(POPh2 )], but ethyl complexes resulted from hydride insertion into the Rh-ethylene bond in the reaction with [Rh(Tp)(C2 H4 )2 ].