Precipitation, amphoterism and ligand exchange

What you’ll learn

• Distinguish deprotonation from ligand substitution.
• Interpret selected transition-ion observations.
• Explain the stability gained with multidentate ligands.

Understand the idea

Adding a base to an aqueous metal ion can remove protons from coordinated water and form a hydroxide precipitate. Familiar observations include pale blue copper(II) hydroxide, green iron(II) hydroxide that darkens on oxidation, orange-brown iron(III) hydroxide, and green chromium(III) hydroxide. A precipitate is a new solid phase; report its colour separately from the surrounding solution.

Amphoteric hydroxides react with both acids and bases. Zinc hydroxide and chromium(III) hydroxide can dissolve in excess hydroxide to form soluble hydroxo complexes. Ammonia can first act as a base and then as a ligand: the pale blue copper precipitate dissolves in excess ammonia to give a deep blue solution. Zinc hydroxide also dissolves in excess ammonia to give a colourless complex; cobalt and nickel hydroxides can form ammine complexes, with appearance depending on oxidation and conditions.

Ligand exchange changes the surrounding ligands without necessarily changing the metal oxidation state. Chloride can replace water in copper(II) or cobalt(II) complexes, often changing both coordination number and colour. Replacing several monodentate ligands by a chelating ligand releases more free particles and can increase system entropy, contributing to greater complex stability. Haemoglobin contains an iron(II) complex with a polydentate ligand; carbon monoxide competes strongly for the oxygen-binding site.

Key formulae

$$\mathrm{[Cu(H_2O)_6]^{2+}+2OH^-\longrightarrow[Cu(H_2O)_4(OH)_2](s)+2H_2O}$$

$$\mathrm{[Cu(H_2O)_6]^{2+}+4NH_3\rightleftharpoons[Cu(NH_3)_4(H_2O)_2]^{2+}+4H_2O}$$

$$\mathrm{Zn(OH)_2+2OH^-\longrightarrow[Zn(OH)_4]^{2-}}$$

$$\mathrm{[Co(H_2O)_6]^{2+}+4Cl^-\rightleftharpoons[CoCl_4]^{2-}+6H_2O}$$

Worked example

Compare copper's oxidation state and coordination number in $[\mathrm{Cu(H_2O)_6}]^{2+}$ and $[\mathrm{Cu(NH_3)_4(H_2O)_2}]^{2+}$.

1. Both water and ammonia are neutral ligands.
2. The overall 2+ charge therefore means copper is +2 in each complex.
3. Both complexes contain six monodentate donor atoms, even though four water ligands have been replaced.

Answer: Copper remains +2 and the coordination number remains 6; this is ligand substitution, not redox.