Zusammenfassung
Methyliminodiacetic acid (H(2)Mida) and imidazole react with copper(II) to form crystals of the square pyramidal complex [Cu(Mida)Im]. One N and two O atoms of the Mida ligand (Cu-N 2.010(1) angstrom, Cu-O 1.955(1) angstrom, and 1.978(1) angstrom) and the imidazole N atom (1.950(1) angstrom) lie at the base of the pyramid. The carboxyl O atom of the neighboring complex lies at the apical position ...
Zusammenfassung
Methyliminodiacetic acid (H(2)Mida) and imidazole react with copper(II) to form crystals of the square pyramidal complex [Cu(Mida)Im]. One N and two O atoms of the Mida ligand (Cu-N 2.010(1) angstrom, Cu-O 1.955(1) angstrom, and 1.978(1) angstrom) and the imidazole N atom (1.950(1) angstrom) lie at the base of the pyramid. The carboxyl O atom of the neighboring complex lies at the apical position (2.411 (1) angstrom); in this way the individual complexes are linked into infinite zigzag chains. Substitution of imidazole by 1,10-phenanthroline gave [Cu-2(Mida)(2)(Phen)H2O]center dot 2H(2)O crystals with two nonequivalent centrosymmetric octahedral anions [Cu(Mida)(2)](2-) of face type (Cu-N 2.023 angstrom and 2.028(2) angstrom,Cu-O-ax 2.579 angstrom and 2.530(2) angstrom,Cu-O-bas 1.952 angstrom and 1.936(2) angstrom). The anions serve as bridges in chains between the [Cu(Phen)H2O](2+) cation fragments to which they are bonded by their axial carboxyl groups. The Cu atom of the cation has a [4+1] environment (with the H2O molecule lying on the axis of the pyramid, and with two N atoms of the ligand and two O atoms of the anions lying at the base).
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