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以四氢呋喃(THF)作溶剂,将[Ru(PPh3)3Cl2]与等摩尔的K[H2B(timMe)2]和过量的苯肼(PhNHNH2)反应,得到1个阳离子配合物[Ru(PPh3){κ3-H,S,S’-H2B(timMe)2}(PhNHNH2)2]Cl·0.5CH2Cl2(1)。单晶X-射线衍射分析显示,配合物1中的中心钌原子与三齿螯合配体[H2B(timMe)2](κ3-H,S,S’)、1个三苯基膦(PPh3)、2个顺式cisPhNHNH2配位形成1个八面体几何构型。通过密度泛函理论计算的价键分析,可确定配合物1中HOMO和LUMO间的能隙较窄,只有3.62 eV。
[Ru (PPh3) 3Cl2] was reacted with equimolar K [H2B (timMe) 2] and excess phenylhydrazine (PhNHNH2) using tetrahydrofuran as solvent to obtain one cationic complex [Ru (PPh3) κ3-H, S, S’-H2B (timMe) 2} (PhNHNH2) 2] Cl0.5CH2Cl2 (1). Single crystal X-ray diffraction analysis showed that the central ruthenium atom in the complex 1 is intimately linked to the tridentate chelating ligand [H2B (timMe) 2] (κ3-H, S, S ’), 1 triphenylphosphine ), Two cis cisPhNHNH2 coordination to form an octahedral geometry. The valence bond analysis calculated by density functional theory confirms that the energy gap between HOMO and LUMO in Complex 1 is relatively narrow, only 3.62 eV.