# Geometry of the transition metal complex in solid dipotassium bis(oxalato) copper(II) tetrahydrate

I was thinking of the geometry of the transition metal complex in solid $$\ce{K2[Cu(ox)2] · 4 H2O}$$ solid around the $$\ce{Cu(II)}$$ ion.

At first I thought square planar, but is it possible or is there any chance that it could be octahedral with two water molecules since it is a hydrate?

Or do brackets always indicate what is directly complexed in this case?

The $$\ce{Cu(II)}$$ atom has square-planar geometry with two bidentate oxalate ligands. Each $$\ce{K+}$$ cation is coordinated by eight oxygen atoms from one bidentate and two unidentate oxalate groups and four water molecules.
Fig. 1. (a) Crystal structures of $$\ce{[K2Cu(ox)2 · 4 H2O]}$$ (1) and (b) $$\ce{[K2Cu(ox)2 · 2 H2O]}$$ (2) with atom numbering scheme. The thermal ellipsoids are drawn at 50% probability.
1. Fan, J.; Sun, W.-Y.; Okamura, T.; Yu, K.-B.; Ueyama, N. The X-Ray Crystal Structural Characterization of Dipotassium Bisoxalato Copper(II) Tetrahydrate, $$\ce{[K2Cu(Ox)2·4 H2O]}$$ $$(\ce{Ox}$$ = oxalate Dianion). Inorganica Chimica Acta 2001, 319 (1), 240–246. DOI: 10.1016/S0020-1693(01)00466-2.