The Electromagnetic Spectrum

The electromagnetic spectrum (EM) spectrum is made up of photons of different wavelengths. Photons, which are unique as they have properties of both waves and particles, enable us to see light and colors through a small portion of the EM spectrum. This visible light portion has wavelengths from roughly 400 to 700 nanometers (a nanometer, “nm,” is 10^-9 meters). Each specific wavelength corresponds to a different color (see Figure 1), and when all the wavelengths are together we see “white light.”

Figure 1

The wavelength and frequency of a wave are inversely proportional--as one gets higher, the other gets lower (this is a consequence of all light traveling at the same speed). Due to this, blue light has a much higher frequency than red light, and consequently has more energy. In general, the farther you go to the right on the EM spectrum, the higher energy the photons have.

Perceiving Color

Color is perceived in two ways, through additive mixing, where different colors are made by combining different colors of light, and through subtractive mixing, where different wavelengths of light are taken out so that the light is no longer pure white. For colors of coordination complexes, we will be talking mainly about subtractive mixing. As shown in Figure 2, the basic idea behind subtractive mixing is this:  white light (which is made from all the colors mixed together) hits an object, and then reacts with it. The object will absorb some of the light, and then reflect and/or (depending on the object) transmit the rest of the light, which goes to our eyes.  We perceive the object as being whatever color is NOT absorbed. In Figure 2, white light (simplified by showing only green, red, and blue) is shown through a solution. The solution absorbs the red and green wavelengths, however the blue light is reflected and passes through, so we see the solution as being blue. Every time we see something as having color, this is the procedure that take place. If none of the light is absorbed, and all is reflected back off, we see the object as being white; if all of the light is absorbed, and there is none left to reflect or transmit through, we see the object as black.

Figure 2

Colors of Coordination Complexes--Crystal Field Splitting

When ligands attach to a transition metal to form a coordination complex, electrons in the d orbital split into high energy and low energy orbitals. The difference in energy of the two levels is referred to as ∆, which is a property both of the metal and the ligands (Figure 3). If ∆ is large, and a lot of energy is required to get electrons into the high energy orbitals, the electrons will instead pair up in the lower energy orbitals, resulting in a complex that is called “low spin” (Figure 4), however if ∆ is small, and it is simple to get electrons into the higher orbitals, the electrons will do so, and remain unpaired (until there are more than five electrons), resulting in a “high spin” complex (Figure 5). Different ligands are associated with either high or low spin--a ligand that is “strong field” will result in a large ∆ and a low spin configuration, while a ligand that is “weak field” will result in a small ∆ and a high spin configuration.  For more details, see Crystal Field Theory (CFT).

Figure 3: d-Orbitals Splitting

                Figure 4: Low Spin, Strong Field (∆_o˃P)                      Figure 5: High Spin, Weak Field (∆_o˂P)

The process of moving an electron to a higher energy orbital, and having it have a higher energy, means that it has the ability to absorb photons of higher energy. When this happens, and certain wavelengths are absorbed, subtractive color mixing takes place and the coordination complex solution becomes colored. If the ions have a noble gas configuration, and have no unpaired electrons, the solutions are seen as colorless (in reality, they still have a measured energy, and absorb certain wavelengths of light, however the wavelengths are not in the visible portion of the EM spectrum so we see no effect).

In general, the larger ∆, the higher energy of photons that is absorbed, and the farther to the right on the EM spectrum (Figure 1) the solution appears. This relationship is described in the equation ∆=hc/λ, where h and c are constants, and λ is the wavelength of light absorbed.

Using a color wheel can be helpful for determining what color a solution will appear based on what wavelengths it absorbs (Figure 6). If a complex absorbs a particular color, it will have the appearance of whatever color is across from it on the wheel. For example, if you determine that a complex absorbs photons in the orange range, you can conclude that the complex will look blue. You can also use this concept, and the equation stated above, to go backwards, and determine ∆ for a complex based on what color it is.

Figure 6

Relating the Colors of Coordination Complexes to the Spectrochemical Series

According to the Crystal Field Theory, ligands that have high spin are considered "weak field" and ligands that have low spin are considered "strong field." How does this relate to the colors we see in a coordination complex? High spin ligands absorb longer wavelengths (lesser frequencies) than do low spin ligands because the splitting energy is greater than the electron pairing energy, thus generating a small splitting. The energy difference, Δ, then determines the color of the coordination complex. According to the spectrochemical series, high spin ligands, which are considered "weak field," will absorb longer wavelengths of light.

Low spin ligands absorb wavelengths of greater frequency because the splitting energy is less than the electron pairing energy, thus generating a greater splitting. Since low spin ligands are considered "strong field," shorter wavelengths of light will be absorbed.  The color that we see is the complementary color of the color absorbed. To predict which possible colors and their corresponding wavelengths are absorbed, we would use the spectrochemical series:

(Strong field/large Δ₀)  CO^-, NO^-, CN^->NO₂^->en>py≈NH₃>EDTA^4->SCN^->H₂O>ONO^->ox^2->OH^->F^->SCN^->Cl^->Br^->I^- (weak field/ small Δ₀)

If a solution appears yellow, what wavelength of light does it absorb? Would you expect it to be low spin or high spin?

Solution

A solution that looks yellow absorbs light that is violet, which is roughly 410 nm. Since it absorbs high energy, the electrons must be raised to a higher level, and ∆ must be high, so you would expect the complex to be low spin.

An octahedral metal complex absorbs light that is 535 nm. What is the crystal field splitting ∆ for the complex? What color is it?

Solution 
In order to solve this question, we need to use the equation ∆ =hc/λ, and the facts that “h,” Planck’s constant, is 6.625E-34 j•s, and “c,” the speed of light, is 2.998E8 m/s.  It is also important to remember that 1 nm is equal to 1E-9 meters. With all this information, the final equation looks like this: ∆ =(6.625E-34 j•s)(2.998E8 m/s)/((535nm)(1m/1E9nm), and ∆ is found to be 3.712 J.
It is not necessary to use any equations to solve the second part of the problem. Light that is 535 nm will be green, and since green light is absorbed, the complex will appear red (refer to Figures 1 and 6 if you need to see where this came from).
*Note: the fact that the complex is octahedral makes no impact when solving this problem. Although the split looks different for complexes of different structures, the mathematics behind it is the same.

You have two solutions, one that is orange, and another that is blue. You know that both solutions are made up of a cobalt complex, however one has chloride ions as ligands, while the other has ammonia ligands. Which solution would you expect to be orange?

Solution
In order to solve this problem, it is necessary to know the relative strengths of the ligands involved. A sample ligand strength list is given here, but see Crystal Field Splitting for a more complete list:  CN^- > en > NH₃ > H₂O > F^- >SCN^- > Cl^-

We can see from this list that NH₃ is a stronger ligand than Cl^-, which means that the complex involving NH₃ will have a greater ∆, and the complex will be low spin. Because of a larger ∆, the electrons will absorb higher energy photons, and the solution will have the appearance of a lower energy color. Since orange light is less energetic than blue light, we would expect the NH₃ containing solution to be orange.

Outside links

• Demonstrations with three primary colored moving lights.

References

1. Clark, Jim. "The Colours of Complex Metal Ions." ChemGuide.Co.Uk. 2003. 3 Dec. 2008 <http://www.chemguide.co.uk/inorganic...ns/colour.html>.
2. Cox, P. A. Instant Notes Inorganic Chemistry. Second ed. Grand Rapids: Garland, Incorporated, 2004.
3. Nassau, Kurt. The Physics and Chemistry of Color : The Fifteen Causes of Color. Second ed. New York: Wiley-Interscience, 2001.
4. Petrucci, Ralph H., William S. Harwood, and Geoff E. Herring. General Chemistry : Principles and Modern Applications. Ninth ed. Upper Saddle River: Prentice Hall PTR, 2006.
5. Petrucci, Ralph H., Carey Bissonnette, F. Geoffrey Herring, Jeffrey D. Madura. General Chemistry: Principles and Modern Applications. Tenth Ed. Upper Saddle River: Pearson Education, Inc. 2011.

Problems

1. What color will a complex be that absorbs light that is 600 nm be?
2. What color will a complex be if it has a ∆ of 3.75E-19 J?
3. Would you expect a violet solution to be high spin or low spin?  What about a red solution?
4. There are two solutions, one which is yellow and another which is violet. The solutions are [Co(H₂O)₆]³⁺and [Co(CN^-)₆]^3-. What are the colors of each solution?

Answers

Highlight for answers

1. Blue. The color absorbed is orange.
2. It is red. Using Δ=hc/λ, h=6.626*10^-34J*s, c=2.998*10⁸m/s, wavelength would equal 530 nm. So green is absorbed, and the complementary color of green is red, so red is the color of the complex.
3. It would be high spin. The complementary color of violet is yellow, which has a wavelength of 570 nm. For a red solution, the complementary color absorbed is green, with a wavelength of 530 nm, so it would be considered low spin.
4. [Co(H₂O)₆]³⁺ is violet and [Co(CN^-)₆]^3- is yellow. Looking at the spectrochemical series, H₂O is a weak field ligand, so it absorbs colors of long wavelengths-in this case, the longer wavelength is yellow, so the color reflected is violet. CN^- is a strong field ligand, so it absorbs colors of shorter wavelengths-in this case, the shorter wavelength is violet, so the color reflected is yellow.

Contributors

• Deyu Wang (UCD)

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