Ligand Substitutions & Kstab

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Ligand Subsitution & K_stab

Complex metal ions have ligands which can be substituted. K_stab is defined as the equilibrium constant for the formation of the complex ion in a solvent from its constituent ions or molecules.

The reaction

The complex ion [ Cu(H₂O)₆ ]²⁺ undergoes a ligand substitution reaction where H₂O is replaced by Cl^-.
The reaction is reversible.
- [ Cu(H₂O)₆ ]²⁺ + 4Cl⁻ $\:\leftrightharpoons\:$ [ CuCl₄ ]^2- + 6H₂O
The equilibrium can be pushed to the right using concentrated hydrochloric acid (HCl).
The reaction can be reversed by adding water. This will push the equilibrium to the left.

[ Cu(H<sub>2</sub>O)<sub>6</sub> ]<sup>2+</sup> vs. [ CuCl<sub>4</sub> ]<sup>2-</sup>

[ Cu(H₂O)₆ ]²⁺ vs. [ CuCl₄ ]^2-

In the reaction, the copper goes from having six water ligands to having four chloride ligands.
- The co-ordination number changes from six to four in the reaction.
- This is because the chloride ions are larger than the water molecules.
- There is only space for four chloride ions around the copper.

Reversible reaction

As this reaction is reversible, we can form an equilibrium expression. This is known as K_stab, or the stability constant.
For the below reaction:
- [ Cu(H₂O)₆ ]²⁺ + 4Cl⁻ $\:\leftrightharpoons\:$ [ CuCl₄ ]^2- + 6H₂O
- K_stab = $\Large\frac {[CuCl_4]}{[Cu(H_2O)_6][Cl^-]^4}$
Looking at the expression, we can see a large K_stab results in the formation of a stable complex ion.

Colour change

The colour change observed should be from blue to yellow.
- However, there is normally a mixture of [ Cu(H₂O)₆ ]²⁺ and [ CuCl₄ ]^2- ions.
- This means the observed colour change is from blue to green (blue + yellow = green).

Substituting Water for Chlorine in a Complex Cobalt Ion

Let's go through another example. The chemistry in this reaction with a complex cobalt ion is exactly the same as the reaction with the with a complex copper ion.

The reaction

The complex ion [ Co(H₂O)₆ ]²⁺ undergoes a ligand substitution reaction where H₂O is replaced by Cl^-.
The reaction is reversible.
- [ Co(H₂O)₆ ]²⁺ + 4Cl⁻ $\:\leftrightharpoons\:$ [ CoCl₄ ]^2- + 6H₂O
The equilibrium can be pushed to the right using concentrated hydrochloric acid (HCl).
The reaction can be reversed by adding water. This will push the equilibrium to the left.

[ Co(H<sub>2</sub>O)<sub>6</sub> ]<sup>2+</sup> vs. [ CoCl<sub>4</sub> ]<sup>2-</sup>

[ Co(H₂O)₆ ]²⁺ vs. [ CoCl₄ ]^2-

In the reaction, the cobalt goes from having six water ligands it to having four chloride ligands.
- The co-ordination number changes from six to four in the reaction.
- This is because the chloride ions are larger than the water molecules.
- There is only space for four chloride ions around the cobalt.

Reversible reaction

As this reaction is reversible, we can form an equilibrium expression. This is known as K_stab, or the stability constant.
For the below reaction:
- [ Co(H₂O)₆ ]²⁺ + 4Cl⁻ $\:\leftrightharpoons\:$ [ CoCl₄ ]^2- + 6H₂O
- K_stab = $\Large\frac {[CoCl_4]}{[Co(H_2O)_6][Cl^-]^4}$
Looking at the expression, we can see a large K_stab results in the formation of a stable complex ion.