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水合双(2-碘乙酸根-κO)(1,10-菲咯啉-κN,N')铜(II)

Aqua-bis(2-iodo-acetato-κO)(1,10-phenanthroline-κN,N')copper(II).

作者信息

Zhao Rengao, Sun Junshan, Lu Jie, Li Jikun

出版信息

Acta Crystallogr Sect E Struct Rep Online. 2009 Jan 31;65(Pt 2):m241. doi: 10.1107/S1600536809002682.

DOI:10.1107/S1600536809002682
PMID:21581830
原文链接:https://pmc.ncbi.nlm.nih.gov/articles/PMC2968381/
Abstract

In the title compound, [Cu(C(2)H(2)IO(2))(2)(C(12)H(8)N(2))(H(2)O)], the Cu(II) ion is coordinated by two N atoms [Cu-N = 2.013 (4) and 2.024 (4) Å] from a 1,10-phenanthroline ligand and three O atoms [Cu-O = 1.940 (4)-2.261 (4) Å] from two carboxyl ligands and a water mol-ecule in a distorted square-pyramidal geometry. One iodo-acetate O atom [Cu-O = 2.775 (4) Å] completes the coordination to form a distorted octa-hedron. Inter-molecular O-H⋯O hydrogen bonds link the mol-ecules into centrosymmetric dimers, which are further packed by π-π inter-actions between the 1,10-phenanthroline ligands into layers parallel to the ab plane. The crystal packing also exhibits short inter-molecular I⋯I contacts of 3.6772 (9) Å and weak C-H⋯O hydrogen bonds.

摘要

在标题化合物[Cu(C₂H₂IO₂)₂(C₁₂H₈N₂)(H₂O)]中,Cu(II)离子由来自1,10 - 菲咯啉配体的两个N原子[Cu - N = 2.013 (4)和2.024 (4) Å]以及来自两个羧基配体和一个水分子的三个O原子[Cu - O = 1.940 (4) - 2.261 (4) Å]配位,形成扭曲的四方锥几何构型。一个碘乙酸根O原子[Cu - O = 2.775 (4) Å]完成配位,形成扭曲的八面体。分子间O - H⋯O氢键将分子连接成中心对称的二聚体,这些二聚体通过1,10 - 菲咯啉配体之间的π - π相互作用进一步堆积成平行于ab平面的层。晶体堆积还表现出3.6772 (9) Å的短分子间I⋯I接触和弱的C - H⋯O氢键。

https://cdn.ncbi.nlm.nih.gov/pmc/blobs/aeca/2968381/4c662e27c6d5/e-65-0m241-fig1.jpg
https://cdn.ncbi.nlm.nih.gov/pmc/blobs/aeca/2968381/4c662e27c6d5/e-65-0m241-fig1.jpg
https://cdn.ncbi.nlm.nih.gov/pmc/blobs/aeca/2968381/4c662e27c6d5/e-65-0m241-fig1.jpg

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本文引用的文献

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