Ligand Exchange - are they reversible?

AI Thread Summary
All ligand exchange reactions are generally considered reversible, meaning that the formation of one complex can be influenced by the concentration of ligands present. In the example provided, where the stability constant (Kstab) for [Cu(H2O)6]2+ is +5 and for CuCl4- is +3, it is possible to form the less stable complex (CuCl4-) by significantly increasing the concentration of chloride ions in the solution. This shift in equilibrium can favor the formation of CuCl4-. However, it is important to note that while the equilibrium can be shifted, the kinetics of the reaction may be slow, leading to situations where a complex with a lower stability constant appears to behave as if it is more stable due to the time required for the reaction to reach equilibrium.
jsmith613
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Are ALL ligand exchange reactions reversible?

Therefore, if the Kstab of one complex is LESS than that of the other complex, would it ONLY be possible to form the LESS stable complex by 'flooding' the mixture with the appropriate ion.

(these values are not real but they illustrate what I mean)
[Cu(H2O)6]2+ - log Kstab - +5
CuCl4- - log Kstab - +3

So to form the CuCl4 I would have to flood a solution of [Cu(H2O)6]2+ with chloride ions to sufficently move the position of eqm to the CuCl4 side

Is this correct
 
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jsmith613 said:
Is this correct

Generally speaking yes, it is just an equilibrium process. Sometimes reaction can be very slow, and even if the stability constants dictate that one ligand should be replaced with the other, it takes so long complex with lower stability constant may behave as a more stable one.
 

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