In aqueous solution, $\ce{Cu^{2+}}$ forms the $\ce{[Cu(H2O)6]^2+}$ complex. Given that water is a stronger ligand than $\ce{Cl-}$, though, why does why does the $\ce{[CuCl4]^2-}$ complex form upon addition of chloride ions?
Answer
[Water ligands] being stronger than $\ce{Cl-}$ ligands will form a $\ce{[Cu(H2O)6]^2+}$ complex.
This is incorrect. Water ligands may be stronger field ligands in the spectrochemical series but that does not in any way affect the strength of the ligand–metal bond or the kinetics and thermodynamics of certain complexes, their formation and their breakdown.
Which ligand forms bonds of which strength to which metal is only very generally determinable; and in general you would always denote binding affinities in a $1:1$ manner: ligand $\ce{X}$ forms strong bonds to metal $\ce{Y}$. So sulphide ions will form strong dative bonds to mercury(II) ions but not-so-strong bonds to titanium(IV) ions. Likewise, palladium(II) if presented with both phosphane and amino ligands will preferentially bind to phosphanes while copper(I) will preferentially bind to amino ligands.
The thermodynamic energy differences between different complexes are often relatively small and typically the kinetic barrier for ligand exchange is low, too. Thus, in many cases which complex is formed depends strongly on the concentrations of individual ions. The mere fact that complexes without aqua ligands can and are formed shows that water is not that good a ligand after all. In fact, due to the positive charge of the central metal having an anionic ligand bind is often preferred due to simple electrostatic interactions.
To answer the actual question you posed: $\ce{[CuCl4]^2-}$ will form because it is the predominant species under a given chloride ion concentration. This can be backrationalised by looking at formation constants $K_1$ to $K_4$. However, aside from quantum chemical calculations there is no way to properly rationalise it a priori.
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