Appendix J: Electronic Absorption Spectroscopy of Metal Complexes

Electronic Structure of d-Metal Complexes. One of the most fascinating aspects of transition metal complexes is the remarkable spectrum of colors which they exhibit. The colors are a result of several features of the complexes which are explicable in reasonably simple terms.

Consider a typical first-transition series metal, e.g., chromium. Free, gaseous chromium atoms have the electronic configuration 1s22s22p63s23p64s13d5, or [Ar]4s13d5. Removal of 3 electrons from this atom (remember that 4s electrons come off first, followed by 3d) gives the CrIII ion with configuration [Ar]3d3. Recall that there are five 3d orbitals, each of which can accommodate a maximum of two electrons (one with "up" spin, one with "down" spin). Since for CrIII there are only three electrons to occupy a total of 10 spots, we say that CrIII has "partially filled" d orbitals (the same can be said of a Cr atom). This characteristic is the hallmark of transition metals and their ions and is the primary reason for the colors exhibited by their complexes.

Many transition metal complexes are octahedral, meaning that the metal ion is surrounded by 6 ligands at the vertices of a regular octahedron. We are going to consider what happens to a transition metal ion in this type of environment. For the sake of simplicity, we will consider the ligands to be capable of sigma bonding only--that is, they each have one orbital which contains a pair of electrons and which can overlap along the internuclear axis with a suitable metal orbital. The resulting complex can be pictured as in Figure 1, where we have introduced a right-handed Cartesian coordinate system in such a way that the ligands lie along the axes, and we have shown the ligand donor orbitals (e.g., if the ligands were NH3 molecules, these orbitals would by sp3 hybrids) pointing in toward the metal, as they must if effective overlap between ligand and metal orbitals is to occur.

How will metal and ligand orbitals interact to form molecular orbitals (MOs) for the complex, and which metal orbitals will be involved in these MOs?

To answer this question we must decide which of the metal ion valence orbitals have appropriate shapes and orientations for overlap with the ligand donor orbitals. Consider first the 4s orbital of the metal. Can it overlap in sigma fashion with any of the ligand orbitals? Since the 4s orbital has equal probability density along each of the 6 Cartesian directions, we conclude that, yes, it can overlap with all of the ligand orbitals in s fashion, as shown in Figure 2.

Similarly we conclude that the 4px orbital can overlap in s fashion with both ligand orbitals which lie along the x axis (but not with ligand orbitals along y or z, since the 4px orbital has no probability density along either of these axes); the 4py orbital can overlap with both ligand orbitals along the y axis; and the 4pz orbital can overlap with both ligand orbitals along the z axis. The picture drawn in Figure 3 represents any one of the three axes:

There is one more point to be made here. You will recall that one lobe of a p orbital has positive sign, while the other lobe has negative sign. Two orbitals which are to overlap effectively must have the same sign, so the ligand orbital which is to overlap with the negative lobe of the p orbital must be negative, as shown above. This will not concern us any more at this point.

So far we have decided that the 4s and the three 4p orbitals of the metal are all suitable for s bonding with the ligands. However, there are six ligands and thus far we have only 4 metal orbitals to use in bonding them. We need a total of 6 metal orbitals to s bond to 6 ligands. We must therefore decide whether any of the 3d orbitals are located suitably for bonding. Let's consider first the 3dz2 orbital, which has positive lobes along the +z and -z directions, and a negative doughnut of probability density in the x-y plane, as shown in Figure 4.

The ligand orbitals lying along the z axis can obviously overlap in å fashion with the two positive lobes; and the ligand orbitals approaching along the x and y axes can overlap in å style with the negative doughnut (if we change their signs). We therefore conclude that the 3dz2 orbital can engage in s bonding. Similar conclusions are reached for the 3dx2-y2 orbital, which has positive lobes along +x and -x and negative lobes along +y and -y.

We now have found 6 metal orbitals which are capable of s bonding to the 6 ligands. But we still have not considered any of the three remaining d orbitals--dxy, dyz, dxz. We will consider one of them. Arguments for the other two are identical. Let's consider dxz, which lies in the x-z plane as shown in Figure 6.

Consider the ligand s orbital which is approaching along the +x axis. Can it overlap in s fashion with dxz? The answer is no. The ligand orbital will overlap with the upper right positive lobe of dxz, but this overlap will be exactly cancelled by an equal and opposite overlap with the lower right negative lobe of dxz. The net overlap is precisely zero. This is equivalent to saying that the dxz orbital has a node along the x axis.

Similar arguments show that none of the four ligands in the xz plane have net overlap with dxz. dyz and dxy fail to overlap with ligand s orbitals for exactly the same reasons. These three orbitals are thus non-bonding in a sigma sense. We conclude that of the originally degenerate (same energy) five 3d orbitals, two are suitable for sigma bonding with ligands and three are not. The d orbitals can therefore be categorized into two groups with respect to their behavior toward s -bonding ligands. Are these two groups still of the same energy?

It must be assumed that if two orbitals, even if intrinsically similar, are differently oriented toward their surroundings, there will be some difference in the energy of an electron depending upon which of them it occupies. Let us see precisely why this difference arises. When each of two atoms has an orbital directed toward the other one (say s 1 and s 2 of Figure 7 below), and these atoms approach closely enough for these orbitals to overlap, the orbitals become "mixed" to form new orbitals, one (the "bonding molecular orbital" or "bonding MO") having relatively high probability density between the two nuclei, and the other (the "antibonding molecular orbital" or "antibonding MO") having relatively low probability density between nuclei. When electrons occupy the bonding orbital, the atoms become bound together, while when electrons occupy the antibonding orbital, a repulsive or antibonding force is set up. Two general statements can be made regarding the two molecular orbitals: 1. the bonding MO always has an energy lower than that of either atomic (or hybrid) orbital from which it is formed, and the antibonding MO always has an energy higher than that of either constituent atomic orbital; 2. If the two

atomic orbitals with which we begin have rather different energies, the resulting bonding MO will have predominantly the characteristics of the lower energy atomic orbital, while the resulting antibonding MO will have predominantly the characteristics of the higher energy atomic orbital. These two statements are illustrated on an energy level diagram below.

Finally, we should state that if two atomic or hybrid orbitals are of such a nature that they have no overlap (such as dxz and the ligand s orbital in Figure 6), there is no mixing and no formation of bonding and antibonding orbitals. The two orbitals are said to be non-bonding with respect to each other.

All of the above considerations apply not only to diatomic molecules, but also to the case of a metal ion interacting in å fashion with six ligands at the apices of an octahedron. Each metal-ligand interaction will give rise to one s bonding and one s antibonding MO. Considering only the interaction between the metal d orbitals and the ligand orbitals, the following energy level diagram results:

The dz2 and dx2-y2 orbitals, which we decided are capable of sigma bonding with the ligands, do so, thus generating two bonding and two antibonding MOs. However, since the metal d orbitals are higher in energy than the ligand sigma orbitals, the antibonding MOs, s *, will have predominantly the character of the metal dz2 and dx2-y2 orbitals, according to general statement 2 above. The dxy,xz,yz metal orbitals, labelled n, which are impervious to s bonding and are therefore non-bonding, are largely unaffected by bond formation and retain their original energies. So instead of maintaining their original five-fold degeneracy in the complex, the d orbitals are split up into a doubly degenerate set, s *, of higher energy and a triply degenerate set, n, of lower energy. This splitting is labelled and is called the ligand field splitting (LFS) since it arises from the "field" generated by the ligands. Now, each ligand s orbital is originally occupied by a pair of electrons. When the complex is formed, these electron pairs will occupy the molecular orbitals of lowest possible energy--that is, the s bonding MOs. Occupation of the ligand s orbitals and the s bonding MOs has been indicated in Figure 10, where represents an electron pair, one electron with up spin and one with down spin. But how about the d electrons originally possessed by the metal ion--which orbitals will they occupy in the complex? Since the degeneracy of the set of d orbitals has been split, electrons no longer will occupy all members of the set with equal probability. Instead, they tend to occupy the more stable ones preferentially, subject to restrictions arising from the Pauli Exclusion Principle and from interelectronic repulsion. (This is taken up more fully in the chapter dealing with magnetism). The electrons will therefore tend to occupy the triply degenerate non-bonding set first.

Let's return to our original example, CrIII. In the complex Cr(NH3)6+3, the chromium d orbitals will be split as above. The three d electrons originally present on chromium will distribute themselves in the available orbitals of lowest energy, as follows:

If we shine light of suitable energy on a molecule, we can cause an electron to move from its present orbital to an orbital of higher energy. The molecule acquires the energy required to do this by absorbing a photon of the light, so that not all of the photons which impinge on, say, a solution of the molecule, will emerge from it. If the photons absorbed have frequencies in the visible region of the electromagnetic spectrum (remember, E = hn =hc/l), the molecule will be colored. Moreover, the color that we see will be due to the photons that are not absorbed by the sample--they pass through it and arrive at our eyes. This color will be the complement of the color which is absorbed by the molecule. If a compound absorbs light of wavelength 550 nm (what is the corresponding frequency? energy?), which is green, it will appear to be reddish-blue, or purple. If a compound absorbs light of wavelength 600 nm (yellow), it will appear to us to be blue. The following color wheel should help you to see this clearly.

Notice that the complement of any color is directly opposite it on the wheel, and that any color can be obtained by mixing the two colors on either side of it. Thus a compound which absorbs red + green (= yellow) will appear blue to the eye. (A compound may absorb two colors when it can undergo two different electronic transitons in the visible region). As it happens, the splitting of the d orbitals (the Ligand Field Splitting, Do) in many transition metal complexes is such that visible light is required to cause the transition of an electron from the triply degenerate lower set to the doubly degenerate upper set, s*. (Such a transition is often called a "d-d" transition, since it involves promotion of an electron from one d orbital to another). Whenever this is the case, as we have seen, the complex is colored. The following figure illustrates the relationship between color and the wavelength range of visible light. Included is a spectrum of the Ti(H2O)6+3 ion, which is purple. As can be seen from the figure, this ion has an absorption band centered at about 550 nm. It therefore absorbs green, transmits blue and red, and looks purple (blue + red).


This is the explanation for the origin of color in transition metal complexes--split d orbitals which are partially filled. The importance of the latter criterion is nicely illustrated by complexes of ZnII, which has the electronic configuration [Ar]3d10. d orbital splitting occurs in ZnII just as it does in the CrIII example; however, since all d orbital spots are already occupied, no transitions within the d orbitals are possible--hence no color is absorbed, and the complexes are white.

References
  1. F.A. Cotton, J. Chem. Ed., 41, 466(1964).
  2. B.H. Mahan and R.J. Myers, University Chemistry, 4th ed., Benjamin/Cummings Publishing Company, Inc., Menlo Park, CA, 1987, Chapter 16.
  3. J.E. Huheey, Inorganic Chemistry, 3rd ed., Harper & Row Publishers, New York, NY, 1983, Chapter 9.

Spectrometer Cells. Cells are containers that are constructed to contain a precise pathlength of solution between two highly polished optical surfaces. Common cell path lengths for electronic absorption spectroscopy are 10 mm (1 cm), 50 mm (5 cm), and 100 mm (10 cm). Most commonly used are the 1 cm cells. Cells are supplied by the manufacturer as a matched set, with a set consisting usually of either 2 or 4 cells. The cells of a set must always be used together and never mixed and matched indiscriminately with other sets of cells. One cell of a set is intended to contain sample + solvent, the other to contain just solvent. It does not matter which cell is used for which.

  1. Handling
    1. Cells are to be handled by the unpolished faces only. Do not touch the polished faces.
    2. Cells are expensive-- > $100 per matched set. When handling them, always do so over a table or bench covered with a towel or cloth, so that if the cell falls from your hand it will not fall far and the blow will be cushioned.
    3. Try to avoid carrying cells directly in your hands. Carry them in their box(es) if at all possible.
  2. Cleaning cells
    1. To clean cell faces, wipe them with lens tissue or with a Kimwipe. NEVER USE A PAPER TOWEL.
    2. If a solvent is necessary, use a Kimwipe wetted with H2O, acetone, or ethanol. Always wipe softly and carefully.
    3. Please do not return a cell to its box until it is completely dry of solvent.
  3. Filling Cells
    1. Place the cell carefully on a towel on the bench top.
    2. Hold the cell with the left hand, fill with a Pasteur pipet to within 5 mm of the top. Insert the Pasteur pipet tip no farther than just below the top of the cell. This will prevent scratching of the polished faces with the pipet.
    3. Stopper the cell, dry the outside of any spillage. You are now ready to measure the spectrum.
  4. Emptying Cells
    1. Pour the contents of the cell either into a beaker or into the original container. From there it can be either discarded in a waste bottle, or saved.
    2. If it is absolutely necessary to remove the contents of the cell using a Pasteur pipet, do so by inserting the pipet so as not to touch the optical faces.
  5. Cleaning cells between runs.
    1. Empty the cell of solution
    2. Rinse the cell a couple of times with the solvent being used. Discard the rinsings.
    3. Rinse the cell with acetone several times.
    4. Aspirate the cell dry.
      • If working in the UV, it is essential to remove all acetone by aspiration, since acetone absorbs in the UV.
      • Never dry a cell by blowing air from the air lines into it, by blowing into it with your mouth, or by waving it around in the air.
  6. Proper choice of cell for the range of interest. There are several types of material used to make cells. These materials differ in the wavelengths of light to which they are transparent. It is important to choose cells of appropriate material for your range. The following list will serve as a guide in choosing the proper cell material. Optical Glass 320-2600 nm
    UV Quartz Glass > 220
    IR Quartz Glass 270-2730
    ES Quartz Glass 200-2000
    Quarasil 200-2600

Sample Preparation for EAS. There are two cases to consider

  1. Extinction coefficients (e) are known.
    1. Calculate the amount of sample required to prepare a convenient volume of solution (usually 5, 10, or 25 mL) based on e values of the absorption bands of the sample. Use Beers Law

      A = ecl

       Substitute the appropriate path length and assume that you want an absorbance of 1.
    2. Weigh the required amount of sample into a volumetric flask of appropriate size. The volume of solution to prepare is largely up to you. However, it is generally desirably to prepare the minimum amount of solution required for the job and consistent with the amount of sample required.
    3. Fill the volumetric flask halfway with solvent, swirl to dissolve the sample.
    4. Fill to the mark with solvent, stopper, and shake well to insure good mixing.
    NOTE: If extinction coefficients are 104 or greater, desired concentration will be 10-4M or less. In this case, direct preparation of 5 or 10-mL volumes of solution will not be possible because the small masses of sample needed cannot be accurately weighed. For example, only 0.0003g of a sample of MW 300 will be required for 10 mL of 1 x 10-4 M solution. This cannot be weighed on a 4-place analytical balance. You should plan to weigh not less than 20 mg of sample in order to insure accuracy in weighing. The procedure to follow in this case is to prepare 5 mL of solution which is 10 times more concentrated than needed. A 1/2-mL aliquot of this, measured by syringe, can then be diluted to 5 mL and the spectrum run. This requires only 9.5 mL of solvent.

  2. Extinction coefficients are unknown. If you do not know where and to what extent your compound absorbs--for instance, if it is a new compound--you will have to obtain the spectrum by preparing solutions according to a systematic trial and error procedure.
    1. Prepare a fairly concentrated solution of the sample, say 0.01-0.05 M, in a 5- or 10-mL volumetric flask. Choose the flask size so that the mass of compound required is no less than 20 mg.

      Ex: For a compound of MW 300, the mass required for 5 mL of a 0.01 M solution is calculated as follows:

      ?g = 0.01 moles/L x 0.005 L x 300 g/mole = 0.0150g = 150 mg.

      Use of a 10-mL volumetric flask would require twice as much and would therefore be wasteful.

      For solutions of transition metal complexes, 0.01 M is a good choice because it will allow you to see d-d bands, which have e values of 1-100 M-1cm-1.

    2. Run the spectrum over the wavelength range 900 to 200 nm.
      1. If the least intense absorption bands have absorbance in the range 0.1 to 1, you have chosen a good starting concentration. Depending on the position of the longest wavelength band, reset the long wavelength limit of the spectrometer to eliminate long stretches where there are no bands, and rerun the spectrum. Pick a convenient limit, like 700 nm, not an inconvenient one, like 721 nm! For example, if the longest wavelenth band has a maximum at 550 nm, it may be appropriate to start the scan at 700 rather than 900 nm. Maintain this new wavelength range for all subsequent dilution spectra.
      2. If the least intense absorption bands are too weak (A < 0.1), prepare a more concentrated solution and rerun until the conditions in 2a are met.
      3. To bring intense absorption bands into the 0-1 absorbance range, dilute the initial solution repeatedly by factors of 2, 5, or 10 to give each band in turn the appropriate intensity. Since absorption bands for a compound can have widely different intensities, several dilutions and corresponding spectral scans may have to be made to bring the absorbances of all bands in turn into the 0-1 range. Be sure to stopper and shake the volumetric flask each time you prepare a dilution, to insure uniformity of concentration. If you don't do this, e values will not be accurate.

        BE SURE TO RECORD EACH DILUTION IN YOUR NOTEBOOK. It is easy to lose track of how many dilutions you have made. Record a dilution as follows: I diluted 0.50 mL of solution 1a to 5 mL using CH2Cl2. New concentration is 1.24 x 10-4 M. I reran the spectrum. See spectrum 3-31-98-2EAS.

         To repeat, once you have selected a final wavelength range in step 2a, do not change it when running dilution spectra.

        A useful guideline is that EAS bands tend to increase in intensity as gets shorter (energy gets higher). The least intense bands will occur in the near IR and Visible regions (900-400 nm), much more intense bands in the UV (< 400 nm).

      4. From the known concentration of the initial solution and the dilution factors, calculate the concentrations of all dilution solutions and, from the measured absorbances, the extinction coefficients of the bands.
    Report data in the format given below:
    CompoundSolventlmax(e), nm(M-1cm-1)
    Co(salen) CH2Cl2490(3060)
    410(14000)
    350(12100)
     NOTE: A good guideline for colored samples (i.e., samples which absorb in the visible region) is that the solution should be concentrated enough so that the color is clearly visible, but of fairly weak intensity in a 1 cm cell. This will usually give absorption bands which fit nicely into the 0-1 absorbance range. However, weaker bands will then be missed.

    For the Near IR and UV regions, trial and error is mandatory.