The ligands in a transition metal complex can be exchanged e.g. ammonia ligands replacing water ligands. This happens either because the incoming ligands can make stronger dative covalent bonds to the central metal ion or simply because the concentration of the new ligand is higher and the equilibrium is displaced.
Replacing water ligands with ammonia ligands
When concentrated aqueous ammonia is added dropwise to a solution of a transition metal ion such as [Cu(H2O)6]2+(aq) we see the formation of a precipitate which may or may not dissolve when excess ammonia is added.
The precipitate is the metal hydroxide. We can explain this in two ways if we remember that ammonia is a base.
- in aqueous solution some ammonia molecules accept a proton from a water molecule to give ammonium ions and hydroxide ions
and so ….
Cu2+(aq) + 2OH–(aq) ⇾ Cu(OH)2(s)
- ammonia removes H+ from two water ligands in the metal aqua complex – the water ligands are now hydroxide ligands (this is more likely if the metal ion has a high charge density and polarises the water ligands)
- the oxidation number of the central metal ion doesn’t change, but the metal ion – hydroxide ion – water ligand complex is now neutral and no longer soluble so it precipitates out
The pale blue copper(II) hydroxide and blue cobalt(II) hydroxide precipitates will dissolve when an excess of ammonia is added, with ammonia ligands replacing the hydroxide ions and some / all of the water ligands. There is no change in the coordination number or charge of the complex as both water and ammonia ligands are neutral.
- copper(II) salts
On addition of concentrated aqueous ammonia we first see a pale blue precipitate which dissolves to form a deep blue solution.
[Cu(H2O)6]2+(aq) + 2NH3(aq) ⇾ [Cu(H2O)4(OH)2](s) + 2NH4+(aq)
[Cu(H2O)4(OH)2](s) + 4NH3(aq) ⇾ [Cu(NH3)4(H2O)2]2+(aq) + 2OH–(aq) + 2H2O(l)
- cobalt(II) salts
On addition of concentrated aqueous ammonia we first see a blue precipitate which dissolves to form a pale yellow solution.
[Co(H2O)6]2+(aq) + 2NH3(aq) ⇾ [Co(H2O)4(OH)2](s) + 2NH4+(aq)
[Co(H2O)4(OH)2](s) + 6NH3(aq) ⇾ [Cu(NH3)6]2+(aq) + 2OH–(aq) + 4H2O(l)
Chromium(III) salts behave in the same way. Addition of concentrated aqueous ammonia to a violet solution of [Cr(H2O)6]3+(aq) gives a grey-green precipitate of Cr(OH)3(s) which dissolves on addition of more ammonia forming a purple solution of [Cr(NH3)6]3+(aq).
[Cr(H2O)6]3+(aq) + 3NH3(aq) ⇾ [Cr(H2O)3(OH)3](s) + 3NH4+(aq)
[Cr(H2O)3(OH)3](s) + 6NH3(aq) ⇾ [Cr(NH3)6]3+(aq) + 3OH–(aq) + 3H2O(l)
We also see metal hydroxide precipitates form when concentrated aqueous ammonia is added to solutions of iron(II), iron(III) and manganese(II) salts. In these three cases, the metal hydroxide precipitates will NOT dissolve on addition of excess ammonia.
- [Fe(H2O)6]2+(aq)
- [Fe(H2O)6]3+(aq)
- [Mn(H2O)6]2+(aq)
Exam tip:
You need to be able to describe all of these reactions, know the colours of the solutions and precipitates and write equations for the formation of the complex ions with state symbols.
Check with your syllabus and text books for guidance on whether you are required to write full complex ion formulae e.g. [Cu(OH)2(H2O)4](s) or the simplified versions e.g. Cu(OH)2(s) in your equations.
Replacing water ligands with chloride ions
When concentrated hydrochloric acid is added dropwise to [Cu(H2O)6]2+(aq), all six water ligands are replaced by four chloride ions. The coordination number of the complex has changed from 6 to 4, the geometry changes from octahedral to tetrahedral and the charge on the complex changes to 2-.
[Cu(H2O)6]2+(aq) + 4Cl–(aq) ⇾ [CuCl4]2-(aq) + 6H2O(l)
Replacing water ligands with polydentate ligands
This is known as chelation as the complexes formed are called chelates. The reactions are always accompanied by an increase in entropy which clearly favours their formation.
[Cu(H2O)6]2+(aq) + EDTA4-(aq) ⇾ [CuEDTA]2-(aq) + 6H2O(l)
Practice questions
- A student carries out a series of test tube reactions on solutions of iron(II) sulfate and chromium(III) chloride.
(a) Identify compounds A – E and describe the observations the student should record in each reaction.
(b) Explain why dilute nitric acid is added the solution of iron(II) sulfate before aqueous barium nitrate is added.
(c) Write ionic equations for the formation of the product in each reaction.
2. The following tests were carried out on a solution containing [CuCl4]2- ions.
(a) Complete the table below. Include all relevant details.
| Test | Observation | Formula of product ion |
|---|---|---|
| (i) dropwise addition of excess water | ||
| (ii) dropwise addition of excess aqueous ammonia, NH3 | ||
| (iii) addition of excess water followed by Na4EDTA(aq) | X X X X X X X X |
(b) Write ionic equations for the reactions observed in (i) and (ii).
Answers
1. (a) A is a gelatinous green precipitate of Fe(OH)2 / Fe(H2O)4(OH)2
B is a white precipitate of BaSO4
C is a grey-green precipitate of Cr(OH)3 / [Cr(H2O)3(OH)3]
D is a violet solution of [Cr(NH3)6]3+
E is a white precipitate of AgCl
(b) Nitric acid is added to ensure that any carbonate ions present are removed from the solution (they bubble off as CO2). BaCO3 is also white precipitate and would obscure observation of the BaSO4 precipitate formed.
(c) There are various correct ionic equations for each reaction (state symbols are compulsory!). You can revise ionic equations here 😊.
A. Fe2+(aq) + 2OH–(aq) ⇾ Fe(OH)2(s) ; [Fe(H2O)6]2+(aq) + 2OH–(aq) ⇾ Fe(OH)2(s) + 6H2O(l) ;
[Fe(H2O)6]2+(aq) + 2OH–(aq) ⇾ [Fe(H2O)4(OH)2](s) + 2H2O(l)
B. Ba2+(aq) + SO42-(aq) ⇾ BaSO4(s)
C. Cr3+(aq) + 3OH–(aq) ⇾ Cr(OH)3(s) ; [Cr(H2O)6]2+(aq) + 3NH3(aq) ⇾ [Cr(H2O)3(OH)3](s) + 3NH4+(aq) ;
[Cr(H2O)6]3+(aq) + 3OH–(aq) ⇾ [Cr(H2O)3(OH)3](s) + 3H2O(l)
D. [Cr(H2O)6]3+(aq) + 6NH3(aq) ⇾ [Cr(NH3)6](aq) + 6H2O(l) ;
[Cr(H2O)3(OH)3](s) + 6NH3(aq) ⇾ [Cr(NH3)6]3+(aq) + 3OH–(aq) + 3H2O(l)
E. Ag+(aq) + Cl–(aq) ⇾ AgCl(s)
2. (a) Be sure to state the colour of the reactant as well as the change to the final product in each case.
| Test | Observation | Formula of product ion |
|---|---|---|
| (i) dropwise addition of excess water | pale yellow solution turns to a green solution* and then a blue solution | [Cu(H2O)6]2+ |
| (ii) dropwise addition of excess aqueous ammonia, NH3 | pale yellow solution turns to a deep blue solution | [Cu(NH3)4(H2O)2]2+ |
| (iii) addition of excess water followed by Na4EDTA(aq) | X X X X X X X X | [Cu(EDTA)]2– |
(b) (i) [CuCl4]2-(aq) + 6H2O(l) ⇾ [Cu(H2O)6]2+(aq) + 4Cl–(aq)
(ii) [CuCl4]2-(aq) + 2H2O(l) + 4NH3(aq) ⇾ [Cu(NH3)4(H2O)2]2+(aq) + 4Cl–(aq)