The chloro complex [IrCl(cod)(dppe)] (dppe = bis-(diphenylphosphane)ethane; cod = 1,5-cyclooctadiene) was found to react with gaseous ammonia, affording the amido-bridged diiridium complex [{Ir(μ-NH 2 )H(dppe)(NH 3 )} 2 ][Cl] 2 (1), whose molecular structure has been solved by X-ray methods. The related cationic complexes [{Ir(μ-NH 2 )H(dppp 4) (dppp = bis(diphenylphosphane)propane; dppb = bis(diphenylphosphane)butane) are only accessible from the reactions of NH 3 (g) with the cationic starting materials [Ir(cod)(dppp)]- [BF 4 ] and [Ir(cod)(dppb)][BF 4 ], respectively. The formation of these species comes from an oxidative addition of an N−H bond of ammonia to the metals. The main structural difference between complex 2 and 3/4 relies on the relative stereochemistry of both ammonia and hydrido ligands; their cisoidal disposition in dppe complex [{Ir(μ-NH 2 )H(dppe)(NH 3 )} 2 ][BF 4 ] 2 (2) directed the reactivity toward dppm (dppm = bis(diphenylphosphane)methane), generating the triply bridged amido complex [{Ir 2 (μ-NH 2 ) 2 (μ-dppm)H 2 (dppe) 2 ]-[BF 4 ] 2 (5). DFT calculations show that the reaction between [IrCl(cod)(dmpe)] (dmpe = bis(dimethylphosphane)ethane) and NH 3 to yield complex [{Ir(μ-NH 2 )H(dmpe)(NH 3 )} 2 ][Cl] 2 comprises four steps: (i) formation of the cationic complex, (ii) replacement of the cod ligand by ammonia molecules, (iii) oxidative addition of the N−H bond to the metal, and (iv) dimerization of the resulting Ir(III) intermediate, step (iii) being the rate-determining one. The calculations reveal that the N−H bond activation takes place heterolytically through an ammonia-assisted stepwise pathway, instead of a concerted homolytic N− H bond cleavage and hydrido formation through a classical three-center transition structure.