Why is k4 Greater than k3 in Cadmium Complex Formation?

[Titan97]

Homework Statement

The formation of cadmium complexes with Br^- exhibits successive equilibrium constants such that log(k₁)=1.56, log(k₂)=0.54, log(k₃)=0.05, log(k₄)=0.37.
Explain why k₄ is greater than k₃

Homework Equations

$$K=e^{\frac{-\Delta H}{RT}}\cdot e^{\frac{\Delta S}{R}}$$

The Attempt at a Solution

After reading chemical equilibrium from peter Atkins's chemical principles, I learned that larger values of K means the reactions is exothermic. (This can be said using the above equation).
Larger value of K, could also mean that $\Delta G$ is strongly negative causing more products at equilibrium.
This explains why some equilibrium constants are large. But this does not explain why K₃<K₄.

 

[Borek]

I would try to analyze symmetry and electronic configuration of complexes involved to see if there are no indications to which one is more stable.

 

[Titan97]

But no information about the structure of complex is given. (I think its tetrahedral)

 

[Borek]

This is even more complicated, as cadmium ion is first complexed by water molecules, they are then replaced by Br^-.

In what context was this question asked? I mean - what is the thing you are learning now?

 

[Titan97]

I am learning chemical equilibrium. But I know the basics about crystal field theory and coordinate compounds. So the reaction is about Br^- replacing OH^- in the complex. Its OK. I would like to learn about this. (I will post the solution given in the solution manual, I don't have it right now).

 

[Titan97]

I got the solution manual @Borek , this is the equation given
$$[CdBr_3(OH)_2]+Br^-\to [CdBr_4]^{2-}+3H_2O$$
According to the author, "there is an increase in the number of particles and hence an increase in dissorderness. This is because of formation of a less restricted coordination sphere."
Am I supposed to learn the equation or is there a general rule to be followed?