Explain cobalt(II) ligand reactions and colour changes
Interpret cobalt(II) reactions with water, ammonia, hydroxide and chloride, distinguishing acid–base precipitation from ligand exchange.
A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 28. Use the first-row transition elements titanium to copper and the species explicitly named in the syllabus. Where redox potentials or stability constants are needed, use the values supplied in the question. Colour descriptions can depend on conditions and concentration. Written practice is self-checked
Before you begin
- Use electron configurations, oxidation numbers, ionic equations, equilibrium expressions and the AS redox and bonding models.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What species gives the familiar pink colour in aqueous cobalt(II)?
What you will learn
- Identify the pink cobalt(II) aqua complex
- Distinguish hydroxide precipitation from excess-ammonia ligand exchange
- Describe chloride exchange and explain how ligand environment changes observed colour
Aqua cobalt(II), hydroxide and ammonia
The pink aqueous cobalt(II) species is represented as [Co(H₂O)₆]²⁺. Adding hydroxide forms cobalt(II) hydroxide: [Co(H₂O)₆]²⁺ + 2OH⁻ → Co(OH)₂(s) + 6H₂O. A small addition of aqueous ammonia can also supply OH⁻ because NH₃ is a weak base, so a hydroxide precipitate may appear first.
With enough excess NH₃, ammonia can act as a ligand and replace coordinated water, forming hexaamminecobalt(II): [Co(H₂O)₆]²⁺ + 6NH₃ ⇌ [Co(NH₃)₆]²⁺ + 6H₂O. The cobalt remains +2 in this ligand-exchange equation. In air, cobalt(II) ammine species can oxidise over time, so an observation after standing may include a redox change as well as ligand exchange.
Chloride exchange produces a different cobalt complex
In sufficiently chloride-rich solution, the pink hexaaquacobalt(II) ion can exchange water ligands for chloride to form blue [CoCl₄]²⁻: [Co(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CoCl₄]²⁻ + 6H₂O.
The equation is an equilibrium, and its position depends on conditions such as chloride concentration and temperature. Colour is evidence that coordination has changed, but mixtures can give intermediate colours.
Pause and explain
What does adding excess chloride to the cobalt(II) aqua equilibrium change?
Worked example
Explain why adding chloride to a pink cobalt(II) aqua solution can produce a blue solution.
Show the worked solution
- Write the aqua/chloro equilibrium.
- Water ligands are replaced by chloride ligands.
- The cobalt oxidation state remains +2 while the ligand environment changes.
Answer [Co(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CoCl₄]²⁻ + 6H₂O. Chloride replaces water around Co(II), producing a blue complex; this is ligand exchange, not oxidation or reduction.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Write the ligand-exchange equation for [Co(H₂O)₆]²⁺ and chloride ions.
Give me a hint
Four chloride ligands replace water while the coordination number remains four in the product.
Compare my reasoning
- Place four Cl⁻ ions on the left.
- Write [CoCl₄]²⁻ on the right.
- Account for the six released water ligands.
[Co(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CoCl₄]²⁻ + 6H₂O.
Look for these in your work
- Stoichiometry balanced
- Complex charge correct
- Water released included
A small amount of aqueous ammonia is added to a pink Co²⁺ solution and a precipitate forms. Give a suitable product and explain ammonia's role.
Give me a hint
Treat aqueous ammonia as a weak base unless conditions support a named ligand complex.
Compare my reasoning
- Ammonia reacts with water to produce some OH⁻.
- OH⁻ reacts with Co²⁺ to form Co(OH)₂(s).
- Do not identify the precipitate as a soluble ammine complex.
Co(OH)₂(s) can form. Aqueous ammonia supplies OH⁻ as a weak base; the observation is hydroxide precipitation, not evidence by itself for an ammine complex.
Look for these in your work
- Base behaviour explained
- Precipitate formula correct
- Ligand-exchange overclaim avoided
With excess NH₃, write the ligand-exchange equation from [Co(H₂O)₆]²⁺ and explain how a solution left in air could later change for a second reason.
Give me a hint
Replace six coordinated water molecules by six NH₃ ligands, then distinguish ligand exchange from oxidation.
Compare my reasoning
- Write [Co(NH₃)₆]²⁺ and release six H₂O molecules.
- The metal remains Co(II) in this exchange equation.
- Prolonged exposure to air can oxidise some Co(II) ammine species to Co(III).
[Co(H₂O)₆]²⁺ + 6NH₃ ⇌ [Co(NH₃)₆]²⁺ + 6H₂O. This exchange retains cobalt in oxidation state +2. If the solution stands in air, oxidation to Co(III) can also occur, so a later colour change should not automatically be assigned only to ligand exchange.
Look for these in your work
- Six-for-six ligand exchange balanced
- Oxidation state retained in exchange
- Later air oxidation distinguished
Common mistakes
- Calling chloride a reducing agent in this colour change
- Ignoring equilibrium and concentration
- Confusing Co(OH)₂ precipitation with a soluble chloro complex
- Saying ammonia never acts as a ligand to Co(II)
- Attributing a colour change after standing in air only to ligand exchange
What can you explain now?
What species gives the familiar pink colour in aqueous cobalt(II)?
Compare with the explanation
[Co(H₂O)₆]²⁺
The pink aqua complex is the starting species in the chloride equilibrium.
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Make it stick
Which cobalt(II) changes are ligand exchange, and which are hydroxide precipitation?
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