Formation of coloured ions (A-level only)

What you'll learn

Why many transition metal ions are coloured.
How d electrons absorb visible light and move to excited states.
Why changing ligand, oxidation state or co-ordination number changes colour.
How a colorimeter can be used to find the concentration of coloured ions.

Colour starts with visible light

Visible light is the part of the electromagnetic spectrum your eyes can detect. White light contains a mixture of wavelengths. A wavelength, given the symbol $λ\lambda$ , is the distance between matching points on neighbouring waves, such as crest to crest.

When light hits a substance, some wavelengths may be absorbed. The wavelengths not absorbed may be transmitted through a solution or reflected from a solid. These remaining wavelengths are what you see as the colour.

Key Idea

Seeing the wavelengths left over

A coloured solution does not look the colour it absorbs. It looks the colour of the light that is transmitted or reflected after absorption.

For example, if a solution absorbs orange-red light strongly, the light reaching your eye may be mainly blue-green, so the solution appears blue-green.

Transition metal ions and d electrons

A transition metal ion often contains a partially filled d subshell. A d subshell contains five d orbitals, and d electrons are electrons in these orbitals.

Many transition metal ions are coloured because their d electrons can absorb visible light and move to a higher energy arrangement.

Definition

Ground state and excited state

The ground state is the lowest-energy arrangement of electrons. An excited state is a higher-energy arrangement formed when an electron absorbs energy.

Not every transition metal ion is coloured. Ions with no d electrons, such as $d^0$ ions, or a completely full d subshell, such as $d^{10}$ ions, are often colourless because the usual d-to-d electron promotion cannot occur.

Common Mistake

Assuming every transition metal ion is coloured

Do not write “all transition metal ions are coloured”. For A-Level, remember that partially filled d orbitals are the important condition for the usual colour explanation.

Why d orbitals split in complex ions

Transition metal ions in solution usually exist as complex ions. A ligand is a species that donates a lone pair of electrons to a central metal ion, forming a co-ordinate bond. Water, ammonia and chloride ions are common ligands.

Definition

Complex ion and co-ordination number

A complex ion contains a central metal ion surrounded by ligands. The co-ordination number is the number of co-ordinate bonds from ligands to the central metal ion.

In an isolated gaseous ion, the five d orbitals have the same energy. We say they are degenerate, meaning equal in energy.

When ligands approach the metal ion, their lone pairs repel electrons in the d orbitals. Some d orbitals are repelled more than others, so the d orbitals split into two different energy levels. The energy gap between these levels is called $\Delta E`.

Splitting of d orbitals in a transition metal complex and promotion of a d electron by absorbed light

Absorbing light: the energy calculation

A d electron can move from the lower d energy level to the higher d energy level if it absorbs a photon with exactly the right energy.

A photon is a packet of electromagnetic radiation. The energy of the photon must match the energy gap $\Delta E`.

ΔE=hν=hcλ\Delta E = h\nu = \frac{hc}{\lambda}

where:

$ΔE\Delta E$ is the energy difference for one electron transition, in J
$h$ is the Planck constant, $6.63 \times 10^{-34}\ \text{J s}$
$ν\nu$ is the frequency, in Hz
$c$ is the speed of light, $3.00 \times 10^8\ \text{m s}^{-1}$
$λ\lambda$ is the wavelength, in m

Example

Calculating the wavelength absorbed

A complex ion has an energy gap of $J\Delta E = 3.20 \times 10^{-19}\ \text{J}$ . Calculate the wavelength of light absorbed.

Choose the correct form of the equation because wavelength is required:
$ΔE=hcλ\Delta E = \frac{hc}{\lambda}$
Rearrange for $λ\lambda$ :
$λ=hcΔE\lambda = \frac{hc}{\Delta E}$
Substitute the values with units:
$J\lambda = \frac{(6.63 \times 10^{-34}\ \text{J s})(3.00 \times 10^8\ \text{m s}^{-1})}{3.20 \times 10^{-19}\ \text{J}}$
Calculate and convert to nm:
$m\lambda = 6.22 \times 10^{-7}\ \text{m}$
This is 622 nm, which is in the visible region.

If a substance absorbs visible light, the colour observed is linked to the wavelengths that are not absorbed.

Tip

Energy and wavelength move oppositely

A larger $ΔE\Delta E$ means a higher-energy photon is absorbed. Since $ΔE=hcλ\Delta E = \frac{hc}{\lambda}$ , larger $ΔE\Delta E$ means a shorter wavelength.

Why the colour changes

The value of $ΔE\Delta E$ is not fixed for a metal ion. It depends on the surroundings of the metal.

Three important changes can alter $ΔE\Delta E$ :

changing the oxidation state of the metal ion
changing the ligand
changing the co-ordination number

The oxidation state is the charge an atom would have if bonding electrons were assigned to the more electronegative atom. In transition metals, different oxidation states often have different colours. For example, vanadium ions in different oxidation states have noticeably different colours.

Changing the ligand also changes the repulsions around the d orbitals. For example, replacing water ligands with ammonia ligands around $Cu2+\text{Cu}^{2+}$ changes the energy gap, so a different wavelength of visible light is absorbed.

Example

Explaining a ligand-substitution colour change

A pale blue aqueous copper(II) complex forms a deep blue complex when excess ammonia is added. Explain why the colour changes.

Identify what has changed around the metal ion: some water ligands have been replaced by ammonia ligands.
Link the ligand change to the d orbitals: ammonia ligands interact differently with the copper(II) ion, so the splitting of the d orbitals changes.
Apply the energy equation: if $ΔE\Delta E$ changes, the absorbed wavelength changes because $ΔE=hcλ\Delta E = \frac{hc}{\lambda}$ .
Connect absorption to observation: a different set of wavelengths is transmitted, so the observed colour changes from pale blue to deep blue.

Common Mistake

Saying the ligand causes the colour directly

The ligand does not simply “give” the ion its colour. The ligand changes $ΔE\Delta E$ , which changes which wavelength is absorbed.

Absorption of visible light in spectroscopy

Spectroscopy is the study of how substances interact with electromagnetic radiation. In visible spectroscopy, you measure which wavelengths of visible light are absorbed.

For transition metal complexes, absorption in the visible region gives evidence about the size of $ΔE\Delta E$ . Different complexes absorb different wavelengths, so they can have different colours.

Common Mistake

Real colours can be more complicated

At A-Level, explain transition metal colours using d electron transitions. Some very intense colours in real chemistry involve other processes, such as charge transfer, but that is beyond the core explanation here.

Using a colorimeter to find concentration

A colorimeter is an instrument that measures how much light is absorbed by a coloured solution.

The key idea is simple: a more concentrated coloured solution contains more coloured ions, so it usually absorbs more light. For dilute solutions, and with the same cuvette path length, absorbance is proportional to concentration.

Definition

Absorbance

Absorbance is a measure of how much light a sample absorbs. Higher absorbance means less light reaches the detector.

Colorimeter setup and calibration graph for determining the concentration of copper(II) ions

A typical method is:

Prepare several standard solutions with known concentrations.
Choose a filter colour that is strongly absorbed by the solution. This is often the complementary colour to the solution.
Use a blank, usually the solvent or reagent mixture without the coloured ion, to zero the colorimeter.
Measure the absorbance of each standard solution.
Plot absorbance against concentration to make a calibration graph.
Measure the absorbance of the unknown solution.
Read the unknown concentration from the graph.

Tip

Good colorimetry technique

Wipe the outside of the cuvette, remove bubbles, keep the same orientation each time, and make sure the solution level is high enough for the light beam.

Example

Finding a copper(II) concentration from absorbance

A calibration graph is a straight line through the origin. A 0.0400 mol dm⁻³ copper(II) standard has absorbance $A = 0.80$ . An unknown copper(II) solution is made by diluting 10.0 cm³ of the original solution to 50.0 cm³. The diluted unknown has absorbance $A = 0.50$ . Find the concentration of the original solution.

Calculate the gradient of the calibration graph:
$mol−1\text{gradient} = \frac{0.80}{0.0400\ \text{mol dm}^{-3}} = 20.0\ \text{dm}^3\ \text{mol}^{-1}$
Use the absorbance of the diluted unknown to find its concentration:
$dm−3\text{concentration} = \frac{0.50}{20.0\ \text{dm}^3\ \text{mol}^{-1}} = 0.0250\ \text{mol dm}^{-3}$
Correct for the dilution. The dilution factor is:
$cm3=5.00\frac{50.0\ \text{cm}^3}{10.0\ \text{cm}^3} = 5.00$
Multiply to find the original concentration:
$dm−30.0250\ \text{mol dm}^{-3} \times 5.00 = 0.125\ \text{mol dm}^{-3}$

Common Mistake

Reading beyond the calibration range

Do not use a calibration graph to estimate concentrations far outside the range of your standards. Dilute the unknown if its absorbance is too high, then remember to apply the dilution factor.

Exam technique

In the exam

Link colour to absorption of visible light, not just “presence of transition metals”.
For colour changes, always mention that ligand, oxidation state or co-ordination number changes $ΔE\Delta E$ , so a different wavelength is absorbed.
In colorimetry questions, describe standards, blanking, measuring absorbance, plotting a calibration graph, and reading off the unknown concentration.

Self review

Check yourself

Why does a transition metal complex appear the colour of the light it does not absorb?
What happens to the absorbed wavelength if $ΔE\Delta E$ becomes larger?
How would you use a colorimeter to find the concentration of an unknown copper(II) solution?

What you'll learn

Why many transition metal ions are coloured.
How d electrons absorb visible light and move to excited states.
Why changing ligand, oxidation state or co-ordination number changes colour.
How a colorimeter can be used to find the concentration of coloured ions.

Colour starts with visible light

Key Idea

Seeing the wavelengths left over

A coloured solution does not look the colour it absorbs. It looks the colour of the light that is transmitted or reflected after absorption.

For example, if a solution absorbs orange-red light strongly, the light reaching your eye may be mainly blue-green, so the solution appears blue-green.

Transition metal ions and d electrons

A transition metal ion often contains a partially filled d subshell. A d subshell contains five d orbitals, and d electrons are electrons in these orbitals.

Many transition metal ions are coloured because their d electrons can absorb visible light and move to a higher energy arrangement.

Definition

Ground state and excited state

The ground state is the lowest-energy arrangement of electrons. An excited state is a higher-energy arrangement formed when an electron absorbs energy.

Common Mistake

Assuming every transition metal ion is coloured

Do not write “all transition metal ions are coloured”. For A-Level, remember that partially filled d orbitals are the important condition for the usual colour explanation.

Why d orbitals split in complex ions

Definition

Complex ion and co-ordination number

A complex ion contains a central metal ion surrounded by ligands. The co-ordination number is the number of co-ordinate bonds from ligands to the central metal ion.

In an isolated gaseous ion, the five d orbitals have the same energy. We say they are degenerate, meaning equal in energy.

Splitting of d orbitals in a transition metal complex and promotion of a d electron by absorbed light

Absorbing light: the energy calculation

A d electron can move from the lower d energy level to the higher d energy level if it absorbs a photon with exactly the right energy.

A photon is a packet of electromagnetic radiation. The energy of the photon must match the energy gap $\Delta E`.

ΔE=hν=hcλ\Delta E = h\nu = \frac{hc}{\lambda}

where:

$ΔE\Delta E$ is the energy difference for one electron transition, in J
$h$ is the Planck constant, $6.63 \times 10^{-34}\ \text{J s}$
$ν\nu$ is the frequency, in Hz
$c$ is the speed of light, $3.00 \times 10^8\ \text{m s}^{-1}$
$λ\lambda$ is the wavelength, in m

Example

Calculating the wavelength absorbed

A complex ion has an energy gap of $J\Delta E = 3.20 \times 10^{-19}\ \text{J}$ . Calculate the wavelength of light absorbed.

Choose the correct form of the equation because wavelength is required:
$ΔE=hcλ\Delta E = \frac{hc}{\lambda}$
Rearrange for $λ\lambda$ :
$λ=hcΔE\lambda = \frac{hc}{\Delta E}$
Substitute the values with units:
$J\lambda = \frac{(6.63 \times 10^{-34}\ \text{J s})(3.00 \times 10^8\ \text{m s}^{-1})}{3.20 \times 10^{-19}\ \text{J}}$
Calculate and convert to nm:
$m\lambda = 6.22 \times 10^{-7}\ \text{m}$
This is 622 nm, which is in the visible region.

If a substance absorbs visible light, the colour observed is linked to the wavelengths that are not absorbed.

Tip

Energy and wavelength move oppositely

A larger $ΔE\Delta E$ means a higher-energy photon is absorbed. Since $ΔE=hcλ\Delta E = \frac{hc}{\lambda}$ , larger $ΔE\Delta E$ means a shorter wavelength.

Why the colour changes

The value of $ΔE\Delta E$ is not fixed for a metal ion. It depends on the surroundings of the metal.

Three important changes can alter $ΔE\Delta E$ :

changing the oxidation state of the metal ion
changing the ligand
changing the co-ordination number

Example

Explaining a ligand-substitution colour change

A pale blue aqueous copper(II) complex forms a deep blue complex when excess ammonia is added. Explain why the colour changes.

Identify what has changed around the metal ion: some water ligands have been replaced by ammonia ligands.
Link the ligand change to the d orbitals: ammonia ligands interact differently with the copper(II) ion, so the splitting of the d orbitals changes.
Apply the energy equation: if $ΔE\Delta E$ changes, the absorbed wavelength changes because $ΔE=hcλ\Delta E = \frac{hc}{\lambda}$ .
Connect absorption to observation: a different set of wavelengths is transmitted, so the observed colour changes from pale blue to deep blue.

Common Mistake

Saying the ligand causes the colour directly

The ligand does not simply “give” the ion its colour. The ligand changes $ΔE\Delta E$ , which changes which wavelength is absorbed.

Absorption of visible light in spectroscopy

Spectroscopy is the study of how substances interact with electromagnetic radiation. In visible spectroscopy, you measure which wavelengths of visible light are absorbed.

For transition metal complexes, absorption in the visible region gives evidence about the size of $ΔE\Delta E$ . Different complexes absorb different wavelengths, so they can have different colours.

Common Mistake

Real colours can be more complicated

Using a colorimeter to find concentration

A colorimeter is an instrument that measures how much light is absorbed by a coloured solution.

Definition

Absorbance

Absorbance is a measure of how much light a sample absorbs. Higher absorbance means less light reaches the detector.

Colorimeter setup and calibration graph for determining the concentration of copper(II) ions

A typical method is:

Prepare several standard solutions with known concentrations.
Choose a filter colour that is strongly absorbed by the solution. This is often the complementary colour to the solution.
Use a blank, usually the solvent or reagent mixture without the coloured ion, to zero the colorimeter.
Measure the absorbance of each standard solution.
Plot absorbance against concentration to make a calibration graph.
Measure the absorbance of the unknown solution.
Read the unknown concentration from the graph.

Tip

Good colorimetry technique

Wipe the outside of the cuvette, remove bubbles, keep the same orientation each time, and make sure the solution level is high enough for the light beam.

Example

Finding a copper(II) concentration from absorbance

Calculate the gradient of the calibration graph:
$mol−1\text{gradient} = \frac{0.80}{0.0400\ \text{mol dm}^{-3}} = 20.0\ \text{dm}^3\ \text{mol}^{-1}$
Use the absorbance of the diluted unknown to find its concentration:
$dm−3\text{concentration} = \frac{0.50}{20.0\ \text{dm}^3\ \text{mol}^{-1}} = 0.0250\ \text{mol dm}^{-3}$
Correct for the dilution. The dilution factor is:
$cm3=5.00\frac{50.0\ \text{cm}^3}{10.0\ \text{cm}^3} = 5.00$
Multiply to find the original concentration:
$dm−30.0250\ \text{mol dm}^{-3} \times 5.00 = 0.125\ \text{mol dm}^{-3}$

Common Mistake

Reading beyond the calibration range

Do not use a calibration graph to estimate concentrations far outside the range of your standards. Dilute the unknown if its absorbance is too high, then remember to apply the dilution factor.

Exam technique

In the exam

Link colour to absorption of visible light, not just “presence of transition metals”.
For colour changes, always mention that ligand, oxidation state or co-ordination number changes $ΔE\Delta E$ , so a different wavelength is absorbed.
In colorimetry questions, describe standards, blanking, measuring absorbance, plotting a calibration graph, and reading off the unknown concentration.

Self review

Check yourself

Why does a transition metal complex appear the colour of the light it does not absorb?
What happens to the absorbed wavelength if $ΔE\Delta E$ becomes larger?
How would you use a colorimeter to find the concentration of an unknown copper(II) solution?

What you'll learn

Colour starts with visible light

Transition metal ions and d electrons

Why d orbitals split in complex ions

Absorbing light: the energy calculation

Why the colour changes

Absorption of visible light in spectroscopy

Using a colorimeter to find concentration

Review Formation of coloured ions (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Formation of coloured ions (A-level only)

What you'll learn

Colour starts with visible light

Transition metal ions and d electrons

Why d orbitals split in complex ions

Absorbing light: the energy calculation

Why the colour changes

Absorption of visible light in spectroscopy

Using a colorimeter to find concentration

Review Formation of coloured ions (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Formation of coloured ions (A-level only)

What you'll learn

Colour starts with visible light

Transition metal ions and d electrons

Why d orbitals split in complex ions

Absorbing light: the energy calculation

Why the colour changes

Absorption of visible light in spectroscopy

Using a colorimeter to find concentration

Review Formation of coloured ions (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Formation of coloured ions (A-level only) Revision Guide

Formation of coloured ions (A-level only)

What you'll learn

Colour starts with visible light

Transition metal ions and d electrons

Why d orbitals split in complex ions

Absorbing light: the energy calculation

Why the colour changes

Absorption of visible light in spectroscopy

Using a colorimeter to find concentration

Review Formation of coloured ions (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Formation of coloured ions (A-level only) Revision Guide