Homo- and Heteropentanuclear Coordination Compounds with Td Symmetry - the Solid State Structures of [MZn4(L)4(L′)6] (M = CoII or Zn; L = chloride or acac; L′=1,2,3-benzotriazolate)

The syntheses of homo and heteropentanuclear coordination compounds with the molecular formulae [MZn4(L)(4)(L')(6)] (M = Co-II or Zn; L = chloride or acac; L' = 1,2,3-benzotriazolate) are reported. These compounds display a highly symmetric coordination unit consisting of a central metal ion (M = Co-II or Zn) which is octahedrally coordinated by 6 tridentate benzotriazolate-type ligands via their N(2) donor atom. The benzotriazolate ligands span the edges of an imaginary tetrahedron thus providing four coordination sites at the corners of the tetrahedron. which are then filled by four zinc ions. The coordination shell of the latter are completed by bidentate acetylacetonate (acac) ligands or by chloride anions, respectively. The solid state structures of two homopentanuclear metal complexes, namely [Zn-5(acac)(4)(bta)(6)]center dot 4C(6)H(12) (1) (acacH = acetylacetone; btaH = 1,2,3-berizotriazole), and [zn(5)Cl(4)(Me(2)bta)(6)]center dot 2DMF (2) (Me(2)btaH = 5,6-dimethyl-1,2,3-benzotriazole) were determined by single crystal X-ray structure analysis. The heteropentanuclear metal complex [(CoZn4Cl4)-Zn-II(Me(2)bta)(6)]center dot 2DMF (3) is isostructural with compound 2. Compound 1 was synthesized from stoichiometric amounts of Zn(acac)(2) and btaH employing dichloromethane as solvent. The synthesis of compound 2 requires addition of an auxiliary base to the DMF solution of anhydrous ZnCl2 and Me(2)btaH. For compound 3 a stoichiometric ratio of Co(NO3)(2)-6H(2)O, anhydrous ZnCl2 and Me(2)btaH was employed during synthesis. Phase purity of all compounds was proved by X-ray powder diffraction (XRPD) analysis, IR spectroscopy, and elemental analysis. Crystal data: for 1 (C80H100N18O8Zn5): monoclinic, space group P2(1)/(c) with a = 23.781(5) angstrom, b = 16.000(3) angstrom, c = 25.170(5) angstrom, beta = 115.29(3)degrees, V = 8659(3) angstrom(3), Z = 4 p = 1.357 g cm(-3). For 2 (C54H62Cl4N20O2Zn5): cubic, space group Fed (3) over barm with a = 23.367(3) angstrom, V = 12759(3) angstrom(3), Z = 8, p = 1.553 g cm(-3). For 3 (C54H62CI4CoN20O2Zn4): cubic, space group Fd (3) over barm with a = 23.443(3) angstrom, V = 12884(3) angstrom(3), Z = 8. p = 1.532 g cm(-3).