Understanding Oxidation States and Electronic Structures of Transition Metals, Study notes of Chemistry

Download Understanding Oxidation States and Electronic Structures of Transition Metals and more Study notes Chemistry in PDF only on Docsity! Lecture Series 10 Transition Metal Chemistry Chapter 18 This is the last material that will be covered on the final. Chapter 18 covers a huge amount of material in a very few number of pages. Not only is there a lot of material in this chapter, the nature of the material is nearly completely new – at least on the surface. There is simply no way that we can cover much of this chapter – nor would we want to! It is simply too much, and I only have a small amount of time in which to lecture on this stuff. Nevertheless, there are a few important topics that we will cover. This chapter concerns itself with the elements that exhibit the chemistry of the d-orbitals. These are called the transition metals. Look for a moment at the fourth row of the periodic table, which contains the elements shown in figure 1.K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 4S 3D 4P Figure 1. The 4th row of the periodic table. The transition metals are the elements that range from Sc (Scandium) to Zn (Zinc).These elements exhibit a range of oxidation states that are unparalleled in the S- and P- elements. We briefly touched on oxidation states in previous sections, but here is a formal definition. The oxidation state of an atom is simply the charge on the atom. For example, Copper can exist as Cu-, Cuo, Cu+, and Cu++. For these atoms and ions, oxidation states are –1, 0, +1, and +2 respectively. Often times students think that oxidation states must have something to do with the element oxygen. This is not true! Similarly, an oxidation process also doesn’t necessarily involve oxygen! If an element is oxidized, then its charge is increased in the positive direction. Examples of oxidation processes include: Cu-
Cu + e Cuo
Cu+ + e Mn+3
Mn+7 + 4e A reduction process is the opposite of an oxidation process. If an element is reduced, then its charge is increased in the negative direction. Any of the above examples of oxidation processes would be reduction processes if they were written in reverse, such as, for the first equation listed, Cu + e
Cu-. This is related, but is not quite the same as ionizing an atom, such as we discussed back in Chapter 15 with the photoelectric effect. In an oxidation/reduction (collectively called redox) process, if an atom or a molecule is oxidized, then something else is reduced, so that charge is balanced in the end. Likewise, if something is reduced, then something else is oxidized. Let’s consider for a moment the electronic structure of Cu, and try to understand these various oxidation states. If we used the Aufbau principle, Hund’s rule, Pauli principle, etc., we would say that the electronic structure of Cuo is 4S2 3D9. Now if we add one more electron, we can get a closed shell, or 4S2 3D10. This would be Cu-, and we would expect it to be fairly stable (it is, after all, a closed shell anion). Now imagine if we removed one electron from Cuo. We would then have 4S2 3D8. This arrangement of electrons doesn’t look especially remarkable – and so one might not expect it to be very stable. However, try moving both of the 4S electrons into the D shell, and you get 4S0 3D10. This is a closed shell cation, and the electronic configuration now does look rather special! In fact, Cu+ is 4S0 3D10, and +1 is a common oxidation state for Cu. Now let’s consider Cu++, which would have an electronic configuration of 4S2 3D7 or of 4S0 3D9, depending on how we think about the +2 oxidation state. Note that here I have been rather casual about moving the 4S electrons back and forth between the 3D orbitals. In fact, this is a fine thing to do, and it is very common for transition metals to grab the S electrons for the d-orbitals when some special stability can be gained. However, for this Cu++ species, it is rather difficult to understand why either of these two electronic configurations would be particularly stable. In fact, Cu++ does exist, but it is not as common as the other oxidation states. From our simple arguments, we would predict that the three most common oxidation states of Cu would be Cu-, Cuo, and Cu+. Cu++ would less common. This is what is observed. The preceding discussion of the oxidation states of copper highlights both the beauty and frustration of inorganic chemistry. It would be great if we could just take the electronic configurations of the transition metal atoms, in their various oxidation states, and predict which ones will to be observed, which ones will be the most stable, etc. The fact is that we can’t do this. However, we can use the electronic configurations as a guide to which oxidation states will be most likely. However, lots of possibilities may be observed – possibilities that we don’t think of when we just look for ‘magic’ electronic configurations. What do we mean by ‘magic’ electronic configurations? Here are a few, with accompanying explanations. The ones highlighted in red are especially stable. 4S0 3D0 rare gas configuration (ex: Sc+2, Ti+4, Mn+7) 4S2 3D0 closed S-shell (ex: Ti(II), V(III), Mn(V)) 4S0 3D5 ½ filled D-shell (ex: Fe+3, Mn+2) 4S1 3D5 6 unpaired, aligned electrons (example Cro, V-, Fe+2) 4S2 3D5 closed S-shell, aligned and unpaired electrons in D shell (ex: Co+2, Ni+3) 4S0 3D10 closed D-Shell (Cuo, Ago, Auo) 4S2 3D10 closed S-shell, closed D-Shell (ex: Hgo, Cdo, Zno) Note that we have, in a few cases, referred to the oxidation states of the various elements using Roman numerals, such as Mn(V). This is equivalent to Mn+5, and is actually a more common way of representing the +5 oxidation state of Manganese. We will use Roman numerals throughout this discussion to represent positive oxidation states of transition metals. z y x x2-y2d z2 dz Figure 3. the other 2 d-orbitals. Note that these are aligned along cartesian axis. :COOC: :C O :C O :CO OC : :CO :CO OC: OC: dx2-y2 :C O :C Od z2 Figure 4. Cr(CO)6. The top image shows the central (green) Cr atom and the 6 carbonyl groups arranged in an octahedral bonding geometry. The bottom image shows how these carbonyl groups point directly toward the two d-orbitals. The second class of orbitals is indicated in Figure 3. Note that, for these two atomic orbitals, the actual lobes of the orbitals lie along the cartesian coordinate axis. Now based on our previous experience with making chemical bonds, our first inclination would be to make a variety of sigma and pi bonds between these d-orbitals and the orbitals of the ligand molecules. We would then get a correlation diagram that contained σ, σ*, π, π*, and nonbonding orbitals. If we wanted to explain transition metal bonding in terms of molecular orbital theory (MO theory), then that is exactly what we would do. In fact, such an approach would eventually work, and we would be able to reproduce the properties of transition metal complexes, although we would have to take MO theory substantially further than we have taken it in this class. For that reason, we aren’t going to go in that direction. Instead, we are going to utilize a different theory, known as crystal field theory. This is that last thing that you will have to know for the final, by the way, and you will just have to know it a little bit. Crystal field theory differs from MO theory in that we don’t exactly ever get as far as making a bond. Instead, we just consider the energies of the atomic d-orbitals on the central transition metal atom under the influence of the surrounding ligands. Consider the molecule Cr(CO)6, which we can call Chromiumhexacarbonyl. Note that the Cr exists as Cro, and that the electronic configuration is 4S1 3D5. Further, note that this structure is octahedral. Each CO looks like :CO, where the lone pair on the carbon is what interacts with the Cr atom. The Chromium atom has one electron in each of the available 5 3d- orbitals. The 6 carbonyl ligands are aligned along the x, y, and z cartesian axis, as shown in figure 4. So now, what is the nature of this chemical interaction? It turns out that the electrons in the d orbitals are repelled by the two electrons in the carbonyl lone pairs. However, not all d-orbitals are affected in the same 0.4 ∆o ∆o 0.6 ∆o Low field splitting dxy dzy dxz ______________________________________________ 0.4 ∆o ∆o 0.6 ∆o High field splitting dxy dzy dxz Figure 5. The two cases of crystal field splitting for the octahedral geometry. way. The d-orbitals that are aligned along the cartesian axis (the x2-y2, and the z2 d- orbitals) experience this repulsion more strongly than do the dxy, dyz, or dxz orbitals. The result is that the energies of the d-orbitals on the Cr atom are split. This is called crystal field splitting. The energy of all 5 d-orbitals are raised relative to their energies in the isolated atom,but the energies of the x2-y2 and the z2 d-orbitals are raised more than the dxy, dyz, or dxz orbitals. The resulting splitting is shown in Figure 18.17 of your text, and I reproduce it here as Figures 5a and 5b. In these figures, I have denoted ∆o as the total energy value of the crystal field splitting. Notice that all the orbitals are raised in energy relative to where they start. In this figure, I show two cases, and these cases correspond to what are classified as weak crystal field splittings, and strong crystal field splittings. What determines whether a crystal field splitting is weak or strong? The nature of the ligand is the most important part. Certain ligands will interact with the d-orbitals only a little bit, and therefore lead to a weak, or low crystal field splitting, while others will interact more strongly. Also note the arrangement of the electrons in the various orbitals. In the low-field splitting case at the top of Figure 5, the most energetically favorable way to arrange the electrons is to keep them all in separate orbitals, with their spins aligned. However, if we split the orbitals by a large amount (bottom of Figure 5), then it becomes more favorable to put all of the electrons into the lowest energy orbitals. Although both of these electronic configurations are paramagnetic, the number of unpaired spins in the low field case is much greater than in the high field case. Thus, if one could measure the number of unpaired electrons (and one can), then this would serve as a measurement for whether the system was in the low- or high- field case. There is another way to get this information, and that is to do spectroscopy. In both of these systems, the lowest energy electronic transitions correspond to taking an electron from one of the 3 lowest energy d-orbitals, and putting it into one of the 2 higher energy d-orbitals. A photon can be used to supply the energy necessary to move these electrons, and a measurement of the photon energy that is required is a direct measurement of ∆o, or the crystal field splitting energy. Obviously, in the high field case, ∆o is greater than it is in the low field case, and so the energy that is required to move this electron from the lower energy d-orbitals to the higher energy d-orbitals is greater for the high field case. It turns out many of these electronic transitions require photons in the visible part of the spectrum, and so it is exactly these electronic transitions that give transition metal complexes their lovely colors, such as Cobalt blue, for example. End of Material for Final Supplementary problems Problem 1: Imagine you have a solution of tetracarbonyldichlorocobalt(III), and that it strongly absorbs light at 700 nm. Now you pour a bunch of ammonia into this same solution in order to make tetraamminodichlorocobalt(III). This new solution absorbs light a 450 nm. What is the electronic configuration of the Co atom? Is it one of the ‘magic’ configurations? What is ∆o for these two complexes? How many unpaired electrons would you expect in each of these complexes? More explicitly, which of these complexes would you expect to have the largest number of unpaired electrons? Problem 2. In order to do magnetic data storage using transition metal complexes, a necessary criterion is for that the transition metals must have a large number of unpaired electrons. The reason is that the unpaired electrons make the complex paramagnetic, and a crystalline solid of those complexes may then be ferromagnetic. Fe(II) compounds (such as FeO) are commonly used for magnetic data storage. Are they high field or low field compounds? Why?