Discuss the stability of coordination compounds.

(N/A) The stability of a complex in a solution refers to the degree of association between two species involved in a state of equilibrium. The stability is expressed by the magnitude of the equilibrium constant (stability or formation constant) for the association.
For a general reaction: $M + nL \rightleftharpoons ML_{n}$
Larger is the stability constant,the higher is the proportion of $ML_{n}$ that exists in a solution.
Free metal ions rarely exist in solution; they are usually surrounded by solvent molecules which compete with the ligand molecules $L$ and are successively replaced by them.
Let us consider the stepwise associations of metal ion and ligands (ignoring solvent molecules):
$M + L \rightleftharpoons ML, K_{1} = \frac{[ML]}{[M][L]}$
$ML + L \rightleftharpoons ML_{2}, K_{2} = \frac{[ML_{2}]}{[ML][L]}$
$ML_{2} + L \rightleftharpoons ML_{3}, K_{3} = \frac{[ML_{3}]}{[ML_{2}][L]}$
$ML_{n-1} + L \rightleftharpoons ML_{n}, K_{n} = \frac{[ML_{n}]}{[ML_{n-1}][L]}$
where $K_{1}, K_{2}, \dots, K_{n}$ are stepwise stability constants. The overall stability constant $(\beta_{n})$ of the formation of species $ML_{n}$ from $M$ and $L$ is given by:
$M + nL \rightleftharpoons ML_{n}$
$\beta_{n} = \frac{[ML_{n}]}{[M][L]^{n}} = K_{1} \times K_{2} \times K_{3} \times \dots \times K_{n}$
Consider the formation of the cuprammonium ion $[Cu(NH_{3})_{4}]^{2+}$,obtained by the stepwise addition of ammonia molecules:
$[Cu(NH_{3})]^{2+}, [Cu(NH_{3})_{2}]^{2+}, [Cu(NH_{3})_{3}]^{2+}$ and $[Cu(NH_{3})_{4}]^{2+}$.
If $K_{1}, K_{2}, K_{3}$ and $K_{4}$ are the stability constants of the successive reactions,then:
$\beta_{4} = \frac{[Cu(NH_{3})_{4}]^{2+}}{[Cu^{2+}][NH_{3}]^{4}}$
The addition of the four amine groups to copper shows a pattern found for most formation constants,where the successive stability constants decrease. In this case,the values are:
$\log_{10} K_{1} = 4.0, \log_{10} K_{2} = 3.2, \log_{10} K_{3} = 2.7, \log_{10} K_{4} = 2.0$
$\log_{10} \beta_{4} = 4.0 + 3.2 + 2.7 + 2.0 = 11.9$,so $\beta_{4} = 10^{11.9}$.
The dissociation constant (instability constant) is the reciprocal of the formation constant.