Anion-Controlled Nuclearity in Nickel Complexes with Potentially Dinucleating, Poly(oxime) Amine Ligands
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- 作者：Elizabeth A. Deters ; Michael J. Goldcamp ; Jeanette A. Krause Bauer ; and Michael J. Baldwin
- 刊名：Inorganic Chemistry
- 出版年：2005
- 出版时间：July 25, 2005
- 年：2005
- 卷：44
- 期：15
- 页码：5222 - 5228
- 全文大小：203K
- 年卷期：v.44,no.15(July 25, 2005)
- ISSN：1520-510X
文摘
Two new ligands consisting of bis(oxime) amine units tethered by a bridge have been synthesized. Their nickelchloride and nickel nitrate complexes have also been synthesized and characterized by X-ray crystallography,FTIR, mass spectrometry, and elemental analysis. One of these ligands, L1 (N,N,N',N'-tetra(1-propan-2-onyl oxime)-diamino-m-xylene), is always dinucleating, while the other ligand, L2 (N,N,N',N'-tetra(1-propan-2-onyl-oxime)-1,3-diaminopropane), shows an unusual anion dependence on the nuclearity. When nickel chloride is used, the ligandacts in a dinucleating manner and coordinates two nickels; however, when nickel nitrate is used, the ligand actsin a monodentate fashion and coordinates only one nickel. Once the mononuclear complex is formed, it is notpossible to add a second nickel if Ni(NO3)2 is used as the nickel source; it is possible, however, to add a secondnickel if NiCl2 is used as the nickel source. The dinuclear complex can be converted to the mononuclear one byeither using silver nitrate to exchange the chloride anions for nitrates or by dissolving the complex in water.Ni2(L1)Cl4(DMF)2.DMF: orthorhombic, P212121, a = 12.2524(11) Å, b = 16.6145(15) Å, c = 20.1234(19) Å, V =4096.5(6) Å3, Z = 4. [Ni2(L2)Cl4(DMF)]2·2DMF: triclinic, P-1, a = 12.5347(5) Å, b = 12.5403(5) Å, c = 14.3504(6) Å, 
= 67.348(1)
, 
= 69.705(1), 
= 81.549(1), V = 1952.25(14) Å3, Z = 1. Ni(L2)·(NO3)2: monoclinic,P21/n, a = 9.6738(3) Å, b = 30.2229(9) Å, c = 15.8238(5) Å, = 97.995(1), V = 4581.4(2) Å3, Z = 8.
= 67.348(1), = 69.705(1), = 81.549(1), V = 1952.25(14) Å3, Z = 1. Ni(L2)·(NO3)2: monoclinic,P21/n, a = 9.6738(3) Å, b = 30.2229(9) Å, c = 15.8238(5) Å, = 97.995(1), V = 4581.4(2) Å3, Z = 8.