Properties and Uses of Transition Metals

Lessons

Senior 6 · Chemistry

Properties and Uses of Transition Metals

5 lessons

Lesson 4 of 5

Why Transition Metal Ions Are Coloured

25 min

Six bottles, six colours

Line up solutions of six transition-metal compounds and you get almost a rainbow. Line up solutions of sodium, magnesium, aluminium, scandium or zinc salts and every one is colourless. Colour is so typical of transition metals that analysts use it to identify them, and it can even measure how much metal is present. To explain it you need to know what "colour" means, and what ligands do to the d orbitals.

Colour is the light that is left

White light is a mixture of all visible wavelengths, from about 380 nm (violet) to 750 nm (red). A coloured solution absorbs some wavelengths; the rest pass through (are transmitted) and reach your eye. What you see is the complementary colour of the light absorbed, the colour opposite it on the colour wheel (Figure 4.1).

Figure 4.1: The colour wheel. The colour you see is the complement of the colour absorbed.

Ligands split the d orbitals

In an isolated gaseous ion the five 3d orbitals all have the same energy: they are degenerate. When six ligands approach along the x, y and z axes to form an octahedral complex, their lone pairs repel the d electrons. Two orbitals, $d_{x^2-y^2}$ and $d_{z^2}$ , point straight at the ligands, so electrons in them are repelled more and their energy rises. The other three, $d_{xy}$ , $d_{xz}$ and $d_{yz}$ , point between the ligands and end up lower. The d orbitals split into two sets separated by an energy gap $\Delta E$ (Figure 4.2).

Figure 4.2: Splitting of the 3d orbitals by six ligands. Absorbing a photon of energy ΔE promotes a d electron to the upper set.

Definition

d–d transition: the promotion of an electron from a lower-energy d orbital to a higher-energy d orbital in a complex, by absorbing a photon whose energy equals the splitting, $\Delta E$ :

$\Delta E = h\nu = \frac{hc}{\lambda}$

where $h = 6.63 \times 10^{-34}\,\mathrm{J\,s}$ is the Planck constant, $ν \nu$ the frequency, $c = 3.00 \times 10^{8}\,\mathrm{m\,s^{-1}}$ and $\lambda$ the wavelength. For most complexes $\Delta E$ matches visible light, so the complex is coloured.

Worked example

Problem. The violet ion $\mathrm{[Ti(H_2O)_6]^{3+}}$ absorbs most strongly at about 500 nm. Calculate the splitting energy $\Delta E$ for one ion and for one mole of ions, and explain the colour. ( $L = 6.02 \times 10^{23}\,\mathrm{mol^{-1}}$ )

Solution.

$\Delta E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{500 \times 10^{-9}} = 3.98 \times 10^{-19}\,\mathrm{J} \text{ per ion}$

$\Delta E \times L = 3.98 \times 10^{-19} \times 6.02 \times 10^{23} = 2.39 \times 10^{5}\,\mathrm{J\,mol^{-1}} = 239\,\mathrm{kJ\,mol^{-1}}$

Answer: 239 kJ mol⁻¹. Light near 500 nm is green to yellow-green. It is absorbed, and the red and blue light that passes through combines to give the violet-purple colour. $\mathrm{Ti^{3+}}$ is $3 d^{1}$ : its single electron makes the jump shown in Figure 4.2.

No partly filled d subshell, no d–d colour

A d–d transition needs an electron in the lower set and a space for it in the upper set. $\mathrm{Sc^{3+}}$ and $\mathrm{Ti^{4+}}$ ( $3 d^{0}$ ) have no d electrons to promote. $\mathrm{Zn^{2+}}$ and $\mathrm{Cu^{+}}$ ( $3d^{10}$ ) have no empty space in the upper set. So their compounds are white and their solutions colourless.

Table 4.1: Colours of some aqueous ions

Ion	d electrons	Colour in water
$\mathrm{Sc^{3+}}$	$d^{0}$	colourless
$\mathrm{Ti^{3+}}$	$d^{1}$	violet
$\mathrm{V^{3+}}$	$d^{2}$	green
$\mathrm{Cr^{3+}}$	$d^{3}$	violet as $\mathrm{[Cr(H_2O)_6]^{3+}}$ ; often green, when other ligands such as sulfate or chloride replace some water
$\mathrm{Mn^{2+}}$	$d^{5}$	very pale pink
$\mathrm{Fe^{2+}}$	$d^{6}$	pale green
$\mathrm{Fe^{3+}}$	$d^{5}$	yellow-brown (from partly hydrolysed ions)
$\mathrm{Co^{2+}}$	$d^{7}$	pink
$\mathrm{Ni^{2+}}$	$d^{8}$	green
$\mathrm{Cu^{2+}}$	$d^{9}$	blue
$\mathrm{Zn^{2+}}$	$d^{10}$	colourless

What changes the colour

The colour depends on the size of $\Delta E$ , which depends on:

the metal and its oxidation state: $\mathrm{Fe^{2+}}$ is pale green and $\mathrm{Fe^{3+}}$ yellow-brown, with the same ligand;
the ligand: some ligands split the d orbitals more than others. In order of increasing splitting: $\mathrm{Cl^-} < \mathrm{H_2O} < \mathrm{NH_3} < \text{en} < \mathrm{CN^-}$ ;
the coordination number and shape: in a tetrahedral complex the splitting is reversed (two orbitals lower, three higher) and much smaller, about $\frac{4}{9}$ of the octahedral value for the same ion and ligands.

Worked example

Problem. $\mathrm{[Cu(H_2O)_6]^{2+}}$ absorbs most strongly at about 800 nm and is pale blue. When excess ammonia is added it becomes $\mathrm{[Cu(NH_3)_4(H_2O)_2]^{2+}}$ , which absorbs most strongly at about 600 nm and is a deep violet-blue. Calculate $\Delta E$ per mole for each and explain the change.

Solution. For one mole, $\Delta E = \frac{hcL}{\lambda}$ , and $hcL = 6.63 \times 10^{-34} \times 3.00 \times 10^{8} \times 6.02 \times 10^{23} = 0.1197\,\mathrm{J\,m\,mol^{-1}}$ .

$\Delta E_{800} = \frac{0.1197}{800 \times 10^{-9}} = 1.50 \times 10^{5}\,\mathrm{J\,mol^{-1}} = 150\,\mathrm{kJ\,mol^{-1}}$

$\Delta E_{600} = \frac{0.1197}{600 \times 10^{-9}} = 2.00 \times 10^{5}\,\mathrm{J\,mol^{-1}} = 200\,\mathrm{kJ\,mol^{-1}}$

Answer: ammonia is higher in the series than water, so it splits the d orbitals more (200 instead of 150 kJ mol⁻¹). The absorption moves to shorter wavelength, from the red end towards orange and yellow, so the transmitted colour shifts from pale blue to deep violet-blue.

Important

Colorimetry. The more concentrated a coloured solution, the more light it absorbs. A colorimeter shines light of the colour the solution absorbs most (for $\mathrm{Cu^{2+}}$ (aq), a red or orange filter) through a sample and measures the absorbance. A calibration graph of absorbance against known concentrations then gives the concentration of an unknown. Pale ions such as $\mathrm{Fe^{3+}}$ are often first turned into an intensely coloured complex, such as the blood-red thiocyanate complex, to make the method more sensitive.

Warning

Absorbed is not seen. A solution that absorbs red light is not red: it is the complementary colour, blue-green. And not every coloured compound owes its colour to a d–d transition: in manganate(VII) and dichromate(VI) the metal has no d electrons ( $d^{0}$ ), and the intense colour comes from electrons moving from oxygen to the metal, a charge-transfer process.

Tip

For a colour question, give four links in the chain: (1) ligands split the d orbitals; (2) a d electron absorbs a photon with $\Delta E = h\nu$ and is promoted; (3) that frequency is removed from white light; (4) the complementary colour is seen. Then say why the ion has a partly filled d subshell.

Chemistry Student: Why do Transition Metals Form Coloured Compounds? (12 min). Watch for the splitting of the d orbitals by ligands, and how the absorbed and complementary colours are linked.

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Practice

Explain why $\mathrm{[Cu(H_2O)_6]^{2+}}$ is coloured but $\mathrm{[Zn(H_2O)_6]^{2+}}$ is colourless.
Which of these ions are coloured in aqueous solution: $\mathrm{Sc^{3+}}$ , $\mathrm{Ti^{3+}}$ , $\mathrm{Cu^{+}}$ , $\mathrm{Fe^{3+}}$ , $\mathrm{Zn^{2+}}$ ? Give a reason for each.
A complex absorbs most strongly at 580 nm. Use Figure 4.1 to predict its colour.
Calculate $\Delta E$ in kJ mol⁻¹ for a complex that absorbs most strongly at 450 nm.
State three factors that affect the size of $\Delta E$ in a complex ion.
Adding concentrated hydrochloric acid to pink $\mathrm{[Co(H_2O)_6]^{2+}}$ gives blue $\mathrm{[CoCl_4]^{2-}}$ . Explain the colour change in terms of the ligand, the shape and the size of $\Delta E$ .
Which colour of filter should a colorimeter use to measure the concentration of copper(II) sulfate solution? Explain.
A student says: "Every coloured compound contains a partly filled d subshell." Use potassium manganate(VII) to evaluate this claim.

Show answersHide answers

$\mathrm{Cu^{2+}}$ is $3 d^{9}$ : the ligands split the d orbitals, and an electron can be promoted from the lower set into the space in the upper set by absorbing visible light. $\mathrm{Zn^{2+}}$ is $3d^{10}$ : the upper set is full, so no d–d transition is possible and no visible light is absorbed.
$\mathrm{Ti^{3+}}$ ( $d^{1}$ ) and $\mathrm{Fe^{3+}}$ ( $d^{5}$ ) are coloured: partly filled d subshells. $\mathrm{Sc^{3+}}$ ( $d^{0}$ ), $\mathrm{Cu^{+}}$ and $\mathrm{Zn^{2+}}$ ( $d^{10}$ ) are colourless.
580 nm is yellow light; the complement is violet.
$\Delta E = \frac{0.1197}{450 \times 10^{-9}} = 2.66 \times 10^{5}\,\mathrm{J\,mol^{-1}} = 266$ kJ mol⁻¹.
Any three: the metal; its oxidation state; the ligand; the coordination number and shape.
Chloride replaces water; $\mathrm{Cl^-}$ splits the d orbitals less than water, and the complex changes from octahedral (six ligands) to tetrahedral (four larger ligands), where the splitting is also smaller. $\Delta E$ falls, the absorption moves to longer wavelength (from blue-green towards orange-red), and the transmitted colour changes from pink to blue.
Red (or orange): the solution absorbs red-orange light most strongly, so the reading changes most with concentration. A blue filter would pass light the solution hardly absorbs.
The claim is false. In $\mathrm{MnO_4^-}$ manganese is +7, which is $3 d^{0}$ , yet the ion is intensely purple. Its colour comes from charge transfer (electrons moving from the oxygen atoms to the manganese), not from a d–d transition. A partly filled d subshell explains the colour of most transition-metal ions, but not all coloured compounds.

Sources and credits

Image: Coloured-transition-metal-solutions.jpg by Benjah-bmm27, via Wikimedia Commons, released into the public domain.
Video: Why do Transition Metals Form Coloured Compounds?, Chemistry Student, embedded from YouTube.
The text, colour wheel, splitting diagram, table, worked examples and Practice items were written and drawn by Ganzaa. Absorption maxima are typical textbook values.

1Six bottles, six colours

2Colour is the light that is left

3Ligands split the d orbitals

4No partly filled d subshell, no d–d colour

5What changes the colour

Six bottles, six colours

Colour is the light that is left

Ligands split the d orbitals

No partly filled d subshell, no d–d colour

What changes the colour