Foundation We know from atomic structure that atoms consist of a positively charged nucleus and various shells filled with electrons. In addition, the periodic table provides us with a basis for orbital filling. Finally, we can describe the relationship between energy and wavelength and detail how it corresponds to colors from our previous study on quantum energy levels.

Goal Transition metal complexes exhibit a wide range of interesting features including coloring and magnetism. We wish to find an explanation for the bonding that occurs with transition metals and relate it to their physical properties.

Observation 1: Those Crafty d electrons Until now, we have only dealt with the s and p orbitals of atoms. What happens when the main property determining electrons are those which reside in the d orbitals? How does this affect such important factors as bonding, oxidation states, and physical properties? Let’s begin by taking a closer look at electron configuration. We know that the first two groups of the periodic table represent the s orbitals. As we move to the transition metals, we now deal with the d-orbitals. Previously, we dealt with L values of 0 and 1 for s and p orbitals. Now, we move up to L =2. Quantum theory tells us that we have our m quantum number ranging from –L to L. As a result, our m values for d-electrons will be -2, -1, 0, 1, and 2, giving us a total of five d-orbitals, which fit up to ten electrons. Since this block of orbitals is further to the right, we know that we should begin by filling s-orbitals and then d-orbitals. Likewise, when we remove electrons they should first be taken from the d-orbitals if common convention is held. We could check this by placing our transition metals in solution and finding out the corresponding oxidation states of the resulting ions. When we do this, we will notice several interesting observations. First, each transition metal has several oxidation states, but nonbonded transition metals only exhibit relatively low oxidation states. Next, many transition metals share the same oxidation states, and each row follows similar patterns. Most striking, however, would be the large range of colors produced by the different ions.

Observed oxidation states of transition metals. The red dots represent the most commonly found states. Picture from: http://content.answers.com/main/content/wp/en-commons/1/19/Transition_metal_oxidation_states_2.png Notice that almost every transition metal has a +2 and +3 oxidation state. Also notice that each row of transition metals seems to follow a similar trend demonstrating adherence to periodic law. This trend is also observed when we look at readily available transition metal compounds. Salts like Pt(NH3)2Cl2 and [Zn(NH3)4]Cl2 exhibit a +2 oxidation state while the salts [Cr(H2O)4Cl2 ]Cl and K3[Fe(CN)6] demonstrate a +3 oxidation state to maintain charge neutrality. Additional complexes akin to [Co(NH3)6]3+ and Pt(en)Cl2 also experience this trend. What can we make of these omnipresent +2 and +3 oxidation states? What do all of these transition metals have in common that would allow this? They all have two s electrons. This could mean that they tend to lose the two s electrons first, and any lost after that are from d-orbitals. This must mean that the s electrons are actually higher in energy once we begin filling the d orbitals. Now, let us look at the patterns. Why do they look like pyramids? As we move across the row, we are decreasing the atomic radii due to increasing the effective nuclear charge. At some point, does this charge start to hold electrons better? This must be where our peaks occur. Once we have made our way halfway across the row, the nuclear charge is considerably large and can have a large attractive effect on the electrons. The shift of progressive rows from the fifth to the sixth atom can likely be explained by the principal quantum number being larger thus leading to a larger radius to start with. Still, two observations elude us. Why do we only see higher oxidation states in bonded complexes, and why do we see such a large spectrum of colors? To answer these questions, we must investigate the nature of bonding when d-electrons are present.

Observation 2: The bonding of Transition Metals Two key characteristics of the bonding nature of transition metals can be gathered from our previous solution experiments. First, we noticed that the transition metals all formed positive ions. As a result, they likely want to bond with species which are negatively charged and can donate electrons. Also, we noticed that transition metals can experience extremely large positive oxidation states, meaning that they likely can bond with several species at once depending on the extent of the oxidation state. We now know that we are looking for electrons donors to bond to our transition metals. This provides us with a general idea of what can be used for bonding. Molecules like H2O, NH3, halogen ions, CN- and ethylenediamine (en) all have at least one lone pair to donate and can serve as ligands.

Representation of ethylenediamine. A lone pair of electrons resides on each nitrogen. Picture from http://www.inchem.org/documents/cicads/cicads/v015ci02.gif Next we must determine how many ligands will add to each transition metal. Judging from the examples given in the previous section, it appears that most transition metal complexes feature four or six bonds. However, we notice one outlier with our Pt(en)Cl2 compound. A similar troubling issue occurs with Ni(en)3. Why would these feature only three bonds when many examples exist where Pt has four bonds and Ni has six bonds? If we examine the structure of our ethylenediamine, we notice that two lone pairs exist. This would allow the en to attach at two sights, which would create a “bridge” and increase our coordination numbers to the expected amounts. Since most t-metal complexes are four or six coordinate, we can use our previous studies of molecular geometry to determine the resulting structures. In four coordinate complexes, tetrahedral and square planar configurations should dominate. When we expand to a six coordinate complex, the structure should become octahedral.

Representations of the molecular geometries of 4 and 6 coordinate compounds. A) The tetrahedral complex features ligands in the interaxial regions. B) Square planar compounds have ligands on the x and y axes. C) Octahedral structures also have z axis ligands. Pictures adapted from http://chemlab.truman.edu/CHEM121Labs/MolecularModeling1.htm Will the types and number of ligands have an effect on the properties of our molecule? Can we use them to explain the color changing and magnetic properties of transition metals?

Observation 3: Ligands Affect the d-orbitals Before we can understand the properties of transition metal complexes, we need to determine how placing ligands on a metal affects the d-orbitals. Whenever a ligand lies along the same axis as a d-orbital, repulsions from the electrons will occur, raising the energy levels of the orbital.

Representations of the d-orbitals. Picture from http://itl.chem.ufl.edu/2045_s00/matter/FG06_023.GIF The octahedral complex has ligands on the x, y, and z axes, so the xy, yz, and xz axes should be equally low in energy since they all exist between the ligand axes allowing them to experience little repulsion. In contrast, the x2-y2 and z2 orbitals lie directly on top of where ligands will go, maximizing repulsion and raising their energy levels resulting in the following pattern:

The square planar complex looks like the octahedral complex without ligands along the z axis. This should result in lowering the energy of any orbital involving the z axis.

Tetrahedral complexes have ligands everywhere that an octahedral complex does not. As a result, we should expect the exact opposite splitting.

Click here for interactive demonstrations: http://wwwchem.uwimona.edu.jm:1104/courses/CFT_Orbs.html

Observation 4: Color Changes Depend on Ligands Let’s test our theory that ligand type affects the color properties of our transition metals. We begin by placing solutions of 20 mL of 1 M Nickel Sulfate in five separate beakers. Then, we add a large amount of water (let’s say 300 mL) to each beaker. Now, we should establish a control group which will remain untouched from here on. The remaining beakers will be used to track color changes. Adding drops of 25% en to each solution should result in a color change. We add 5 mL to the second beaker, 10 mL to the third beaker, 15 mL to the fourth beaker and 20 mL to the fifth beaker, giving us the resulting solutions:

Picture and experiment adapted from http://genchem.chem.wisc.edu/demonstrations/Gen_Chem_Pages/20transmetalpage/transmetalmain.htm The fourth and fifth solutions are practically the same. Adding more should hopefully drive the reaction further. Additional drops have no effect, so it can be determined that the reaction is complete. Originally, we began with a green color when the only substances present were nickel and water. As we added en, the color changed. Each new color must correspond to a new amount of en bonding to each nickel. Nickel tends to form six coordinate compounds, so we can assume that the first complex must be Ni(H2O)6 2+. Each successive compound must be some combination of water and en bonded to Ni. Since en has two lone pairs, it actually bonds at two different spots to the Ni making a bridge. This means we can expect one en to take the place of two waters. In order to maintain the six coordinate nature of Ni, our other beakers must contain Ni(H2O)4(en)2 2+ , Ni(H2O)2(en)2 2+, and Ni(en)3 2+

The four compounds formed in our experiment. Picture from www.chemlearn.chem.indiana.edu/demos/Changing.htm From our studies of quantum energy levels, we know that light can be emitted corresponding to the difference in energy levels. E=hc/λ, so large energy differences should correspond to smaller wavelengths and purple colors while small energy differences should result in large wavelengths and colors closer to red.

The color spectrum. Small wavelengths (high energy) correspond to violet, while large wavelengths (low energy) result in red. Picture adapted from: http://askabiologist.asu.edu/research/seecolor/images/cones_color_graph2.gif Where will these transitions occur? We could expect them to come from the d-orbitals because they are not involved in bonding since they do not overlap with the s and p orbitals of the ligands. In the case of nickel, we have an octahedral complex with a +2 oxidation state and eight d-electrons giving us the resulting filling:

When light hits our solution, it must promote an electron from the lower lying xy, xz or yz orbitals to any of the higher unpaired orbitals. Do our color results make sense with this model? We would expect en to have a stronger electric field than that of water, so we would expect it to create a larger energy gap and colors closer to red. This is not what happened. Instead, we saw colors coming closer and closer to violet. How could this possibly be reconciled? We can perceive colors for two reasons. Either we see it because that color is the only color not absorbed or because all colors of visible light are absorbed except for a particular color known as its complimentary color. An artist’s color wheel demonstrates the phenomenon of complimentary colors.

Representation of the color wheel. Transition metals appear as the compliment of the color that they absorb. Picture adapted from: http://www.cs.berkeley.edu/~sequin/CS184/IMGS/Colorwheel.GIF Does this fit better with our observation? Originally we saw green when we expected a small energy gap and large wavelength. Red is the compliment of green and has a large wavelength. We then perceived blue followed by violet when we expected an increase in energy gaps and decreases in wavelength. Blue’s compliment in orange, and violet’s compliment is yellow. This pattern demonstrates decreasing wavelengths as expected, so we can conclude that the color we see is a result of absorption of complimentary colors. Additional color experiments can be done with Cr(III) to get a relative idea of the strengths of the electric fields of common ligands.

As we move down the table, we notice a decrease in the wavelength of the complimentary color, meaning our energy gap is increasing. We can now make a general ranking of the strengths of the electric fields given off by ligands. F- , H2O , NH3, CN- (least to greatest)

Observation 5: Magnetism Another interesting characteristic of transition metals is their magnetic ability. Since the last electrons reside in the d-orbitals, this magnetism must be due to having unpaired d electrons. Complexes of Fe can be used to demonstrate the properties of magnetism in transition metals. Fe prefers to exist as Fe3+ and is known to have a coordination number of six. Since the configuration of Fe3+ has five d electrons, we would expect to see five unpaired spins in complexes with Fe. When we submit [FeF6]3- to a magnetic susceptibility balance, we find the five unpaired spins as expected, but when we run the same test on [Fe(CN)6]3- we only get one unpaired electron making it a weaker magnet. Perhaps this can be explained similar to our color changes. We expect CN- to have a stronger electric field than that of F- , so the energy level differences should be greater for the cyanide complex resulting in something like this

In order for this to make sense, there must be some sort of energy benefit to having paired spins for our cyanide complex. At this point, the energy level difference must be more than the repulsive energy of pairing electrons together. Since systems strive to achieve the lowest energy possible, the electrons will pair up before moving to the higher orbitals. This is referred to as low spin, while waiting to pair electrons is known as high spin. Now, we must determine if this holds true for tetrahedral and square planar complexes. Tetrahedral complexes have naturally weaker splitting because none of the ligands lie within the plane of the orbitals. As a result, we would not expect to see a need to pair electrons early, but we may not need to worry about this due to the nature of tetrahedral compounds. Several common examples include [CoCl4]2-, [V(NH3)6]3+, and [Zn(NH3)4]2+. All of these either have too many or too few d-electrons to warrant worrying about high or low spin. Finally, we have to figure out how square planar complexes react to splitting. Square planar compounds stem solely from transition metals with eight d electrons, so a magnetic susceptibility test will clearly demonstrate if pairing occurs. [Ni(CN)4]2-, [Pt(NH3)3Cl]+, and [PtCl4]2- are all diamagnetic, meaning all electrons are paired. Since this encompasses the full spectrum of ligand strength, we can conclude that square planar compounds are always low spin.

Representation of orbital pairing in square planar complexes. These compounds are always low spin. For an interactive demonstration, click here: http://wwwchem.uwimona.edu.jm:1104/courses/dOrbs.html

Review and Discussion Questions Rank the following complexes in order of number of unpaired electrons from least to greatest: [V(CN)6]3-, [Co(CN)6]3-, [FeF6]3-, [Fe(CN)4]2-, and [Zn(en)2]2+. Which coordination complex is the most likely to be yellow: [FeCl6]3-, [Fe(CN)6]3-, [Fe(H2O)6] 3-, or [Fe(NH3)6]3-? Why? A student gave the following response on a chemistry exam: It is possible for four-coordinate complexes of the same transition metal to exhibit both tetrahedral and square planar configurations. Two such examples are [PdF4]2- and [Pd(CN)4]2-,. The fluorinated compound will likely be square planar due to the small size of the F ligands allowing for everything to fit in the same plane. This will give us a diamagnetic compound. The cyanide compound will be tetrahedral and have only two unpaired electrons due to the high spin inducing nature of cyanide. Assess the accuracy of the preceding statement and make any necessary corrections. Based on the information given in figure one, predict which two elements in the first row of transition metals are not classified as transition metals. Provide reasoning for your answers (how do they differ). A company has asked you to design a new magnet. Provide both a good example and a bad example of a transition metal to use and explain why you made these selections. Insert Solution Text Here

References and Additional Resources 1) D. Oxtoby, H. Gillis, and N. Nachtrieb. Principles of Modern Chemistry 5th edition. Thompson Brooks-Cole. 2002. pp. 635-665 2) J. McMurry and R. Fay Chemistry. 4th edition. Piarson Prentice-Hall. 2004. pp. 863-9143) T. Brown, H. Lemay, and B. Bursten. Chemistry: The Central Science. 10th edition. Prentice Hall. 20064) For more demonstrations click here http://genchem.chem.wisc.edu/demonstrations/Gen_Chem_Pages/20transmetalpage/transmetalmain.htm