Regulation of Geometry around the Ruthenium Center of Bis(2-pyridinecarboxylato) Complexes by the Nitrosyl Moiety: Syntheses, Structures, and Theoretical Studies
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cis-[Ru(NO)Cl(pyca)2] (pyca = 2-pyridinecarboxylato), in which the two pyridyl nitrogen atoms of the two pycaligands coordinate at the trans position to each other and the two carboxylic oxygen atoms at the trans position tothe nitrosyl ligand and the chloro ligand, respectively (type I shown as in Chart 1), reacted with NaOCH3 to generatecis-[Ru(NO)(OCH3)(pyca)2] (type I). The geometry of this complex was confirmed to be the same as the startingcomplex by X-ray crystallography: C13.5H13N3O6.5Ru; monoclinic, P21/n; a = 8.120(1), b = 16.650(1), c = 11.510(1) Å; = 99.07(1); V = 1536.7(2) Å3; Z = 4. The cis-trans geometrical change reaction occurred in thereactions of cis-[Ru(NO)(OCH3)(pyca)2] (type I) in water and alcohol (ROH, R = CH3, C2H5) to form [{trans-Ru(NO)(pyca)2}2(H3O2)]+ (type V) and trans-[Ru(NO)(OR)(pyca)2] (type V). The reactions of the trans-form complexes,trans-[Ru(NO)(H2O)(pyca)2]+ (type V) and trans-[Ru(NO)(OCH3)(pyca)2] (type V), with Cl- in hydrochloric acid solutionafforded the cis-form complex, cis-[Ru(NO)Cl(pyca)2] (type I). The favorable geometry of [Ru(NO)X(pyca)2]n+ dependedon the nature of the coexisting ligand X. This conclusion was confirmed by theoretical, synthetic, and structuralstudies. The mono-pyca-containing nitrosylruthenium complex (C2H5)4N[Ru(NO)Cl3(pyca)] was synthesized by thereaction of [Ru(NO)Cl5]2- with Hpyca and characterized by X-ray structural analysis: C14H24N3O3Cl3Ru; triclinic,P, a = 7.631(1), b = 9.669(1), c = 13.627(1) Å; = 83.05(2), = 82.23(1), = 81.94(1); V = 981.1(1) Å3;Z = 2. The type II complex of cis-[Ru(NO)Cl(pyca)2] was synthesized by the reaction of [Ru(NO)Cl3(pyca)]- or[Ru(NO)Cl5]2- with Hpyca and isolated by column chromatography. The structure was determined by X-ray structuralanalysis: C12H8N3O5ClRu; monoclinic, P21/n; a = 10.010(1), b = 13.280(1), c = 11.335(1) Å; = 113.45(1); V= 1382.4(2) Å3; Z = 4.

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