Abstract

Abstract The reaction of the electrophilic transition metal iridium with tellurium, indium, and bromine resulted in black, shiny crystals with the composition [Ir2Te14Br12]2(InBr4)2. X‐ray diffraction on single‐crystals revealed a triclinic structure (space group P$\bar{1}$) that contains two crystallographically distinct, centrosymmetric clusters, (Te10)[Ir(TeBr3)2]2+ and (Te10)[Ir(TeBr4)(TeBr2)]2+, as well as two tetrahedral InBr4– anions per unit cell. The center of the positively charged cluster is a Te10·– radical anion. This biconvex tricyclo[5.1.1.13, 5] unit consists of two angulated Te4 rings that are linked by two almost linear μ‐Te bridges. The radical causes an intense single ESR signal at g = 1.999. Molecular DFT‐calculations show that the unpaired electron populates an orbital of the Te10‐unit. Computational variation of the number of electrons causes noticeable variations in the Te–Te distances, providing strong evidence for the Te10·– radical. The decatellurium unit coordinates two iridium(III) cations as a bridging bis‐tetradentate ligand. Two terminal bromidotellurate(II) groups complete the slightly distorted octahedral coordination of each transition metal atom. The two [IrTe6] polyhedra share a common edge. The constitutions of the terminal ligands differ, including only TeBr3– anions in one type of clusters, but a combination of TeBr42– and TeBr2 groups in the other. The coordinating tellurium atoms of the central Te10·– and of the terminal groups act as electron pair donors, thereby fulfilling the 18‐electron‐rule for the iridium(III) cations.

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