Ion Colours

Factors affecting the colour of complex ions

To explain why transition metal complexes have different colours, you'll need to be familiar with the concept of the electromagnetic spectrum. Perhaps you’ve heard of the idea that visible light, or ‘white light’, is made up of a mixture of colours. You can see this if you shine white light through a prism - it splits into all of the colours of the rainbow. The light we can see is just a small part of the electromagnetic spectrum, which includes other kinds of light we cannot see like infra-red, and ultraviolet.

Fig. 2 - The visible spectrum

Each of the light colours in the visible spectrum has a certain wavelength. At one end we have red light, which has a wavelength of 700 nanometers (nm). At the other end we have violet light, with a wavelength of 400 nm. All the other colour waves exist between red and violet and have specific wavelength values between 700 and 400 nm, e.g. 438.4, or 605.0. As you can see, there's an infinite range of colours in the visible spectrum!

When white light passes through a transition metal complex in an aqueous solution, it absorbs some of the wavelengths. The remaining light waves are reflected or transmitted. These remaining light waves give the solution its characteristic colour. For example, copper(II) sulphate solution appears cyan (a pale blue) because, as light passes through it, it absorbs the light in the red region of the visible spectrum. In other words, the light that comes out the other side of the solution has all the other light wavelengths except for red. Mixing these wavelengths together forms the colour cyan.

Fig. 3 - Complex ions in solution absorb certain wavelengths of light and transmit others

Of course, this is a rather simple explanation for the colours of complex ions. We can explain the colours of aqueous solutions of complex ions even further by looking at the arrangement of the electrons in their 3d energy level.

Degenerate and non-degenerate d-orbitals

In the sub-section below, we shall be exploring electron promotion. In order to do this, however, we need to explore what degenerate and non-degenerate d-orbitals are.

In a transition element metal or ion there are 5 d-orbitals. Without any ligands bonded, all 5 are at the same level, therefore, they can be described as degenerate. However, after ligands bind to the transition metal ion, it is no longer by itself. When ligands bind, they form dative covalent bonds and this splits the 5 d-orbitals into two sets. These two sets do not have equal energy and are therefore described as non-degenerate. Now let's explore how the degenerate 5 d-orbitals split in an octahedral complex and a tetrahedral complex.

Splitting in an octahedral complex:

Splitting in a tetrahedral complex:

In a tetrahedral complex the ligands line up with dyz, dxz and dxy, meaning that there is more repulsion than at dx²-y² and dz². Therefore, dyz, dxz and dxy will be at a higher energy level, while dx²-y² and dz², will be at a lower and more stable energy level compared to the former. As in the splitting in octahedral complexes, the energy difference between the two is ΔE.

Electron promotion

A transition metal is one that forms stable ions with a partially filled d subshell. How do we know the colours of complex ions in solution are connected to their partially filled d subshell? Well, when we observe the colour of non-transition metal ions in solution, they are colourless. For example, scandium and zinc, although in the d-block, do not form ions with a partially filled d orbital. Their ions appear clear in solution. On the other hand, the transition metal ions with an incomplete 3d energy level form coloured solutions.

Just how does an incomplete 3d energy level give complex ions in solution colour? Let's take copper(II) sulphate and zinc sulphate as an example. As we've discussed, copper(II) sulphate solution is cyan blue, while zinc sulphate solution is colourless. They both contain sulphate, so any difference in colour must be caused by the copper and zinc ions. Cu²⁺ has the electron configuration 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁹ whilst Zn²⁺ has the electron configuration 1s², 2s², 2p⁶, 3s², 3p⁶, 3d¹⁰. You'll notice that Cu²⁺ has a partially-filled 3d subshell whereas Zn²⁺ has a complete 3d subshell.

In an aqueous solution, water ligands attach to the copper ion and split its 3d orbitals into two sets of non-degenerate orbitals. This means the two sets have slightly different energies, as explained above. The lower energy level has six electrons while the higher one has three. We call this splitting. Splitting only occurs if the 3d subshell is partially filled.

Fig. 5 - Ligands split the 3d orbitals in Cu²⁺ into two sets of non-degenerate orbitals

When the copper(II) sulphate solution absorbs light from the visible spectrum, an electron from the lower energy level gets promoted or excited by moving to the higher energy level. We say it moves from the ground state to an excited state. The energy absorbed by the excited electron depends on the difference in energy between the two levels.

Fig. 6 - An excited electron gets promoted when it absorbs a specific light frequency

A small energy difference between the split 3d energy levels of the Cu²⁺ ion means that the excited electron absorbs a small light frequency. Light frequencies are inversely proportional to their wavelengths, so large light frequencies have small wavelengths, and vice versa. In other words, red light has a large wavelength and a small frequency but violet light has a small wavelength and a large frequency. In our example, the promoted electron in the Cu²⁺ ion absorbs a small light frequency. This means that it absorbs light with a large wavelength, namely light from the red end of the visible spectrum. That is why we see copper(II) sulphate solution as cyan blue, the complementary colour of red.

On the other hand, the zinc ion has a completely filled 3d subshell so no splitting takes place. This means the electrons don't get excited. There aren't transitions that absorb light energy, so zinc complexes are colourless.

Fig. 7 - The Zn²⁺ ion has a completely filled 3d sub-shell, so no splitting takes place between the 3d orbitals

How to find ∆E of an excited electron

As we've discovered, when an electron moves from the ground state to an excited state it absorbs energy in the form of light. Light travels in waves and particles we call photons. Electrons absorb light energy in the form of photons. Albert Einstein showed us that the amount of energy in a photon equals the frequency of a lightwave, v, multiplied by Planck’s constant (6.626 x 10^-34 Js), h. In other words, E_photon = hv.

We express the change in energy when a electron goes from the ground state to an excited state as ∆E. You can find the wavelength of light absorbed by the excited electron by using the equation:

$∆\mathrm{E}=\mathrm{hv}=\frac{\mathrm{hc}}{𝝀}$

Where:

• ∆E is the exact amount of energy absorbed in joules, J

• h is Planck’s constant, 6.626 x 10^-34 Js

• v is the frequency of light in hertz, Hz, or s^-1

• c is speed of light, 3.0 x 10⁸ ms^-1

• 𝝀 is the wavelength of light in meters, m

Why do these ions change colour?

You have seen that the amount of light energy absorbed (∆E) by electrons in the split d orbital gives ion complexes their unique colours. Any factor that changes the level of splitting in the d orbitals changes ∆E and so changes the colour of the ion complex. These include:

• Oxidation state
• Ligands
• Coordination number

Let's discuss how these factors change the ion colours.

Oxidation state

The oxidation state or oxidation number of an ion is the same as its ionic charge. An ion with a 2+ charge has an oxidation state of +2 and an ion with a 3- charge has an oxidation state of -3. Transition metals have a unique quality that allows them to have varying oxidation states. As oxidation state increases, so does splitting in the d orbitals and vice versa. Increased splitting simply means there is a larger energy gap between the split orbitals. This means that when there is a change in oxidation state, the colour of the ion complex also changes.

Ligands

Atoms or molecules bonded to a complex ion are called ligands. The identity of the ligands bonded to a complex ion can also change its colour. How so? Well, there are five 3d orbitals that all have the same energy. You can say they are degenerate. When ligands attach to a transition metal ion by dative covalent bonds, the 3d orbitals split into two sets of non-degenerate orbitals. That is to say, they no longer have the same energy - some have a higher energy than the others.

Electrical fields surrounding the ligands influence the energy gap between the split 3d orbitals. The stronger the electrical field, the larger the energy gap. The difference in energy between the split 3d orbitals determines the size of the light wavelength an ion complex can absorb. For example, ammonia causes more splitting than water in copper(II) sulphate. Water ligands give copper(II) sulphate a light blue colour, while ammonia ligands give it a deep navy colour.

Fig. 8 - Copper(II) sulphate in water and in ammonia

Coordination number

The coordination number is the number of ligands attached to the central ion. It is usually a number between two and nine.

A change in coordination number changes the colour of an ion by changing the amount of splitting in the 3d orbitals. For example, an octahedral ion has more splitting in the 3d orbitals than a tetrahedral ion. This change in coordination number changes the colour of the ion.

To illustrate this, have a look at the reaction between copper(II) sulphate solution and concentrated hydrochloric acid shown below. When we add the acid slowly and continuously to the ion complex, the colour changes gradually from blue to green to yellow.

$[\mathrm{Cu}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6{]}^{2+}+4{\mathrm{Cl}}^{-}\rightleftharpoons {\left[{\mathrm{CuCl}}_4\right]}^{2-}+6{\mathrm{H}}_2\mathrm{O}$

From the equation, can you see if the ligands have changed? What about copper's coordination number?

• The 6 water ligands have changed to 4 chloride ion ligands.

• The coordination number has changed from 6 to 4.

• Copper's oxidation state is still the same even though the overall charge on the ionic molecule has changed from 2+ to 2-. This is because the copper ion itself still has a charge of 2+.

Remember that these factors all affect the splitting of the 3d orbitals, which changes the colour of the complex ion.

Visible light spectroscopy

Spectroscopy is the name of a branch of chemistry that studies the absorption and emission of light and other kinds of radiation. Different substances absorb different wavelengths of light. You have seen this in the case of copper(II) sulphate, which absorbs red light from the visible spectrum and transmits blue light. So copper(II) sulphate appears blue. Even colourless substances absorb specific wavelengths, but in the ultraviolet region, so our eyes don't notice. In visible light spectroscopy, we can use this knowledge to identify substances by means of a colorimeter.

A simple colorimeter measures light absorbance, or the amount of visible light that a substance absorbs. The amount of light absorption depends on the concentration of the substance in solution. This means a colorimeter can help to determine the concentration of coloured ions in solution.

In visible light spectroscopy, you pass different frequencies of light through a coloured filter and a sample of complex ions in solution. A detector and recorder is placed at one end to record the frequency of the transmitted light. The filter you use must match the part of the spectrum that the coloured complex absorbs the most. For example, a solution that absorbs red light appears blue, so you must use a red filter. This way, only red light passes through the solution, and maximum absorption can take place.

Fig. 9 - A colorimetry setup

You will need to produce a calibration graph or calibration curve to figure out the concentration of the sample solution. Let us use the example below to find the concentration of a sample of [Cu(H₂O)₆]²⁺ ions in solution.