When PdCl2 is in aqueous or alcoholic solution, it has four chloride ligands, provided they are available. This is how the wacker works.I have a book on oxidations using palladium, and it describes the entire method of the oxidation. The reactive complex is formed by displacement of one of the four chloride ligands with the alkene pi system. Then, another chloride ligand is displaced by a solvent molecule, followed by deprotonation. This complex then rearranges to a sigma system and oxidatively decomposes.
No, no. This is not how the Wacker oxidation works. There are three steps in the catalysis cycle, Oxidative addition, Crossmetallation, and Reductive elimination. When you have your PdCl2, it oxidativly adds the alkene, then a -OH to the metal center. Next, it couples the -OH to the alkene, adds a molecule of water, and gives off a mol of HCl. Then it reductivly eliminates the alcohol as the ketone and another mol of HCl. You are left with Pd0, which is re-oxidised by the CuCl2, which in turn is re-oxidised by the HCl released, forming a mini cycle.
I didn't make myself clear earlier about chlorine. There is only two ways that Chlorine will be bidentate or polydentate. Chlorine will form very weak coordination complexes with very electropositive atoms. Alkali metal ions, Hydrogen, early transitional metals, lower main group elements such as Sn, etc. The other is in hypervalent complexes, like ClF3 or ClO4. In order to form hypervalent species, the ligands have to be more electronegative than Chlorine itself.