Complex Ions, Ligands, Ligand Exchange

Learn and apply Complex Ions, Ligands, Ligand Exchange in the published Chemistry course sequence.

  • GCE A-Level H2 Chemistry 9476-2027
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Complex Ions, Ligands and Ligand Exchange: Orientation

This lesson follows the exact 9476 ligand-exchange sequence: define ligand and complex, track the copper(II) complexes formed with water, ammonia and chloride ions, and explain oxygen/carbon monoxide competition in haemoglobin.

Use the Transition Elements hub for the lesson sequence and d-Orbital Splitting and Colour for the orbital explanation behind an exchange colour change.

Definitions (Must Know)

A. Ligand

A ligand is an ion or molecule that donates a lone pair of electrons to a central metal ion to form a coordinate bond.

B. Complex

A complex contains a central metal atom or ion surrounded by ligands joined through coordinate bonds. A charged complex is written in square brackets with its overall charge outside, for example [Cu(H₂O)₆]²⁺.

Detailed Explanations

A. Why a complex forms

A ligand supplies both electrons in the new bond. Its lone pair is donated into an available orbital on the electron-deficient metal ion, producing a coordinate bond.

B. Water–ammonia sequence for copper(II)

Adding a little ammonia first supplies enough OH⁻ for a pale-blue copper(II) hydroxide precipitate: [Cu(H₂O)₆]²⁺ + 2OH⁻ → Cu(OH)₂(s) + 6H₂O

In excess ammonia, ligand exchange produces the deep-blue ammine complex: [Cu(H₂O)₆]²⁺ + 4NH₃ ⇌ [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O

Describe the observations in order: pale-blue precipitate first, then dissolution to a deep-blue solution in excess ammonia.

C. Water–chloride ligand exchange

[Cu(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CuCl₄]²⁻ + 6H₂O

Increasing chloride concentration shifts the equilibrium towards the chloride complex. Dilution shifts it back towards the aqua complex. State that a different ligand environment changes the d-orbital splitting and hence the colour.

D. Oxygen–carbon monoxide exchange in haemoglobin

Oxygen binds reversibly as a ligand at the metal centre. Carbon monoxide competes for the same site and forms a more stable complex, displacing oxygen and reducing the number of sites available for oxygen transport. No detailed biological mechanism is required.

E. Equation audit

For every exchange equation, check the central metal, each ligand, total atoms and overall charge. Brackets identify the complete complex; the charge belongs outside them.

Worked Examples

Modelled example 1

Oxidation State in a Cyanide Complex

Core

Problem

Calculate the oxidation state of iron in [Fe(CN)₆]⁴⁻.
Study the worked solution
  1. Total ligand charge

    Method

    Assign six CN⁻ ligands a total charge of -6.

    Reason

    Each cyanide ligand carries charge -1.

    Working

    6(-1) = -6.
  2. Match the complex charge

    Method

    Solve x-6 = -4.

    Reason

    Metal plus ligand charges equal the overall complex charge.

    Working

    x = +2.

Common misconception 2

Charge of a Cobalt Ammine Complex

Find and correct the mistake

Learner claim

A learner writes [Co(NH₃)₆]⁸⁺ by adding six to the cobalt charge. Correct the formula.

Account for ligand charge

NH₃ ligand charge
Complex charge

View solution step by step
  1. Assign ligand charge

    Method

    Treat all six ammonia ligands as neutral.

    Reason

    Ligand count does not imply ionic charge.

    Working

    6(0) = 0.
  2. Write the complex

    Method

    Keep the cobalt(II) charge.

    Reason

    + 2 + 0 = +2.

    Working

    [Co(NH₃)₆]²⁺.

Challenge 3

Copper(II) Hydroxide from a Hexaaqua Ion

Minimal support

Equation transfer

Write the equation for [Cu(H₂O)₆]²⁺ forming a precipitate with hydroxide ions.

Balance charge, hydroxide and water

OH⁻ coefficient
Released waters

Hints

Hint 1: solid
The precipitate is Cu(OH)₂.
Hint 2: ligands
Account for all six water ligands on the product side.
View solution step by step
  1. Form the neutral solid

    Method

    Combine copper(II) with two hydroxides.

    Reason

    Cu(OH)₂ is charge neutral.

    Working

    Cu²⁺ + 2OH⁻ → Cu(OH)₂(s).
  2. Account for aqua ligands

    Method

    Release six water molecules.

    Reason

    The starting copper species is hexaaqua.

    Working

    [Cu(H₂O)₆]²⁺ + 2OH⁻ → Cu(OH)₂(s) + 6H₂O.

Mind Stretchers

Mind stretcher 1: Following a reversible chloride exchangeExtension

Question. Concentrated chloride solution changes a pale-blue copper(II) solution to a different colour. Adding much water restores the pale-blue colour. Explain both changes and state what evidence would distinguish this process from a redox reaction.

Show Hint

Write the reversible exchange equation, then compare the copper oxidation state before and after the colour change.

Show Answer

Added chloride shifts [Cu(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CuCl₄]²⁻ + 6H₂O towards the chloride complex. Dilution lowers chloride concentration and shifts the equilibrium back towards pale-blue [Cu(H₂O)₆]²⁺. The copper oxidation state remains +2 on both sides; showing no oxidation-number change distinguishes ligand exchange from redox.

Mind stretcher 2: Competing ligands in haemoglobinExtension

Question. Carbon monoxide binds more strongly than oxygen to the same metal centre in haemoglobin. Explain, using ligand exchange and equilibrium, why even a modest carbon monoxide exposure can reduce oxygen transport.

Show Hint

Treat oxygen and carbon monoxide as competing ligands at the same binding site, not as redox reagents.

Show Answer

Oxygen and carbon monoxide act as ligands competing for the same metal centre. Stronger binding by carbon monoxide shifts ligand exchange towards the carbon monoxide complex, so fewer sites remain available for reversible oxygen binding. Oxygen transport therefore falls even though carbon monoxide is not itself consuming the oxygen in a redox reaction.