Treating SbCl\sb3, SnCl\sb4.5H\sb2O, or Pb(OAc)\sb2.3H\sb2O with NaCo(CO)\sb4 forms the open complexes Sb{Co(CO)\sb4}\sb3, Sn{Co(CO)\sb4}\sb4, and Pb{Co(CO)\sb4}\sb4, respectively. The unstable antimony complex was determined to be Sb{Co(CO)\sb4}\sb3 by the similarity of its infrared spectrum with that of known Bi{Co(CO)\sb4}\sb3. Isostructural Sn{Co(CO)\sb4}\sb4 and Pb{Co(CO)\sb4}\sb4 were characterized by single crystal X-ray diffraction. Each contains a Group 14 element tetrahedrally surrounded by four trigonal bipyramidal Co(CO)\sb4 groups. Heating solutions of Bi{Co(CO)\sb4}\sb3 forms BiCo\sb3(CO)\sb9, which contains a closed cobalt triangle capped by a bismuth atom. A bridging carbonyl lies along each Co-Co bond. The thermal decomposition is reversible, and a kinetic study of the carbonylation of BiCo\sb3(CO)\sb9 indicates this transformation to be first-order with respect to (BiCo\sb3(CO)\sb9) and P(CO) in n-hexane (500-900 psi, 50-65 C), with an activation energy of 68+/$-2kJ.mol\sp{-1}$. Reducing Bi{Co(CO)\sb4}\sb3 forms [Bi{Co(CO)\sb4}\sb4]\sp−. This complex is isostructural to Sn{Co(CO)\sb4}\sb4 and Pb{Co(CO)\sb4}\sb4, but it is not isoelectronic, as the bismuth is a hypervalent 10-electron center. Reducing BiCo\sb3(CO)\sb9 produces (Bi\sb2Co\sb4(CO)\sb11) \sp− and (Co(CO)\sb4]\sp−. Single crystal X-ray analysis (Cp\sb2Co\sp+ salt) determined the core structure of the bismuth anion to consist of a Bi\sb2Co\sb2 tetrahedron with Co(CO)\sb3 units capping the two Bi\sb2Co triangular faces. X-ray analyses of (PPN) (Sb\sb2Co\sb4(CO)\sb11) and (PPN) \sb2 (Sb\sb2Co\sb4(CO)\sb11) showed these complexes to have nearly identical frameworks, which are isostructural and isoelectronic to their bismuth homologues. These complexes are electron rich and do not conform to conventional bonding formalisms. Extended Huckel calculations agree with observed structural changes that upon reduction of the monoanion, the added electron enters an antibonding orbital primarily localized between two cobalt atoms bridged by a carbonyl. Treatment of BiCo\sb3(CO)\sb9 with PPh\sb3 gives Bi{Co(CO)\sb3PPh\sb3}\sb3. Bi{Co(CO)\sb4}\sb3,BiCo\sb3(CO)\sb9, and Sb{Co(CO)\sb4}\sb3 all react with PhC\sb2Ph through an apparent radical mechanism to form Co\sb2(CO)\sb6{PhC\sb2Ph} as the only infrared observable product.