The reaction of iridium powder with an excess of selenium and SeBr₄ yielded lustrous, vermillion crystals of the mononuclear iridium complex [IrBr₃(SeBr₂)₃]. The transition metal is coordinated octahedrally by three SeBr₂ and three bromide ligands with facial or meridional configuration. Three different modifications were obtained under similar conditions: a-fac-IrBr₃(SeBr₂)₃, space group P\bar{1}, with a = 789.4(1) pm, b = 830.4(1) pm, c = 1334.4(1) pm, α = 81.634(5)°, β = 84.948(5)°, γ = 67.616(4)°; m-fac-IrBr₃(SeBr₂)₃, space group P2₁/n, with a = 1205.3(1) pm, b = 962.4(1) pm, c = 1383.9(1) pm, β = 91.114(3)°; mer-IrBr₃(SeBr₂)₃, space group P2₁/n with a = 859.7(1) pm, b = 1284.3(1) pm, c = 1437.5(1) pm, β = 94.427(3)°. A lower bromine content in the starting composition resulted in shiny, deep-red crystals of [Se₉(IrBr₃)₂]. X-ray diffraction on a single-crystal revealed a tetragonal lattice (space group I4₁/a) with a = 1245.4(1) pm and c = 2486.8(1) pm at 296(1) K. In the [Se₉(IrBr₃)₂] complex, a crown-shaped uncharged Se₉ ring coordinates two iridium(III) cations as a bridging bis-tridentate ligand. Three terminal bromide ions complete the distorted octahedral coordination of each transition metal atom.