Titanium complexes with fluoride ligands

The i5p-titanium(IV) atom is hard, ie, not very polarizable, and can be expected to form its most stable complexes with hard ligands, eg, fluoride, chloride, oxygen, and nitrogen. Soft or relatively polarizable ligands containing second- and third-row elements or multiple bonds should give less stable complexes. The stabihty depends on the coordination number of titanium, on whether the ligand is mono- or polydentate, and on the mechanism of the reaction used to measure stabihty. 

With the monodentate fluoride ion, it is difficult to have two H2O ligands in trails allowing condensation of opposed coplanar edges. This mode of condensation is possible only with a bidentate ligand ([HSO4]) which leaves only one water molecule in the coordination sphere. As a result, only anatase can form. In both cases, this mechanism may only take place if the complexes are sufficiently stable. Equilibria between various species are probably involved. Under these conditions, it is difficult to know what is the precursor of the solid. Since the oxide always contains some amount of sulfate difficult to remove, it is reasonable to speculate that the complexes are rather stable and that the formation of the solid takes place by incorporation of the sulfated titanium complexes by olation. This is also probably the case with the fluoride. Therefore, the complexing ions of titanium... 

Since thermodynamic data are not available, the electronegativity-pH diagram (Figure 5.18) shows that complexation of titanium by sulfate and fluoride ions may take place iti a strongly acid medium. It cannot take place with chloride ions. The zero-charge complex precursor is [Ti(OH3)X(OH2) ]° (X = [HS04] , F ), with n = 2 or 1 if. the ligand is bi- or monodentate respectively. 

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