The reaction of Ir-3(CO)(9)(mu(3)-Bi) with PhAu(NHC) (1), where NHC = 1,3-bis(2,6-diisopropylphenylimidazol-2-ylidene), has yielded the compound Ir-3(CO)(8)(Ph)-(mu(3)-Bi)[mu-Au(NHC)] (2) by the loss of one CO ligand and the oxidative addition of the Au-C (phenyl) bond of 1 to one of the iridium atoms. The Au(NHC) group bridges one of the Ir-Bi bonds of the cluster. On the basis of X-ray crystal structural analysis and molecular orbital and quantum theory of atoms in molecules calculations, the Au-Bi interaction was determined to be substantial and is comparable in character to the Ir-Bi and Ir-Ir bonds in this cluster. Compound 2 reacts with 2 equiv of HSnPh3 to yield the compound Ir-3(CO)(7)(SnPh3)(2)(mu(3)-Bi) [mu-Au(NHC)] (mu-H) (3), which contains two terminally coordinated SnPh3 ligands. Compound 3 reacts with H2O to yield the compound Ir-3(mu(3)-Bi)(CO)(7)[mu-Ph2Sn(OH)SnPh2][mu-Au(NHC)] (4) by cleavage of a phenyl ring from each of the SnPh3 ligands and formation of a bridging OH group between the two tin atoms to form a chelating Ph2Sn(OH)SnPh2 ligand.