Synthesis and structure of a binuclear ruthenium 4-chlorobenzamidato complex

Abstract

The compound Ru₂Cl(4-Cl-C₆H₄CONH)₄ was prepared by reaction of Ru₂Cl(O₂CCH₃)₄ with 4-Cl-C₆H₄CONH₂ at 180°C. Crystals of the composition Ru₂Cl(4-Cl-C₆H₄CONH)₄CH₃OH were obtained by slow diffusion of CH₃OH containing Et₄NCl into a Me₂SO solution of the compound. The structure of the crystalline product, which loses solvent of crystallization on removal from the mother liquor, was solved by X-ray crystallography by mounting a single crystal in a capillary containing the mother liquor. The crystals belong to the space group P1^- (triclinic crystal system) with a = 12.731(3) Å, b = 14.389(3) Å, c = 12.604(3) Å, α = 103.41(2)°, β = 106.43(2)°, γ = 64.90(2)°, V = 1988.6(8) ų and Z = 2. There are two half ruthenium dimers linked by a Cl atom and an uncoordinated solvent CH₃OH molecule per asymmetric unit. The ruthenium dimers lie on two centers of inversion at 0, 0, 0 and 1/2, 0, 0. The chloride ions bridge dinuclear cations in the crystal, forming infinite zigzag chains. The average Ru-Ru distance is 2.296[1] Å and each ruthenium atom has a RuClN₂O₂ coordination sphere where the average Ru'-Ru-Cl angle is virtually linear (175.68[6]°). The metal oxidation states in the complex are + 2 and + 3, giving an average value of + 2.5. The arrangement of four bridging 4-Cl-benzamidato ligands is of the 2 : 2 type. The average Ru-N, Ru-O, Ru-Cl distances and Ru(1)-Cl(1)-Ru(2) angle are 2.036[6] Å, 2.044[5] Å, 2.583[2] Å and 117.26(8)°, respectively. The IR spectrum of the compound shows two N-H stretches at 3380 and 3340 cm^-1. The electronic spectrum of the compound in Me₂SO exhibits bands at 558 nm (ε = 340 M^-1 cm^-1), 425 nm (1000) and 320 nm (22,700).