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What is the coordination geometry and oxidation state of [Nd(C2O4)3]3- complex and how does it compare to the [La(C2O4)3]3- complex?

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The coordination geometry and oxidation state of the [Nd C2O4 3]3- complex can be determined by examining the individual components of the complex.Nd is a neodymium ion, which typically has an oxidation state of +3. C2O4 is an oxalate ion  C2O4 2-, which acts as a bidentate ligand, meaning it can form two coordinate bonds with the central metal ion. In the [Nd C2O4 3]3- complex, there are three oxalate ligands, each forming two coordinate bonds with the Nd3+ ion, resulting in a total of six coordinate bonds.The coordination geometry of the [Nd C2O4 3]3- complex is octahedral, as there are six coordinate bonds arranged around the central Nd3+ ion.Now let's compare this to the [La C2O4 3]3- complex. La is a lanthanum ion, which also has an oxidation state of +3. The complex has the same oxalate ligands  C2O4 2- acting as bidentate ligands. There are three oxalate ligands, each forming two coordinate bonds with the La3+ ion, resulting in a total of six coordinate bonds.The coordination geometry of the [La C2O4 3]3- complex is also octahedral, as there are six coordinate bonds arranged around the central La3+ ion.In conclusion, both the [Nd C2O4 3]3- and [La C2O4 3]3- complexes have an oxidation state of +3 for the central metal ion and an octahedral coordination geometry. The main difference between the two complexes is the central metal ion  Nd3+ in the first complex and La3+ in the second complex , which may result in slightly different chemical and physical properties due to the differences in size and electronic configuration of the metal ions.
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