K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr4S 3D 4PFigure 1. The 4th row of the periodic table. The transition metals are the elements thatrange from Sc (Scandium) to Zn (Zinc).Lecture Series 10 Transition Metal Chemistry Chapter 18This 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. Notonly is there a lot of material in this chapter, the nature of the material is nearlycompletely new – at least on the surface. There is simply no way that we can cover muchof this chapter – nor would we want to! It is simply too much, and I only have two daysin which to lecture on this stuff. Nevertheless, there are a few important topics that wewill 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 theperiodic table, which contains the elements shown in figure 1.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 aformal definition. The oxidation state of an atom is simply the charge on the atom. Forexample, 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 theelement oxygen. This is not true! Similarly, an oxidation process also doesn’tnecessarily involve oxygen! If an element is oxidized, then its charge is increased in thepositive direction. Examples of oxidation processes include:Cu- à Cu + eCuo à Cu+ + eMn+3 à Mn+7 + 4eA 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 ofoxidation processes would be reduction processes if they were written in reverse, such as,for the first equation listed, Cu + e à Cu-.This is related, but is not quite the same as ionizing an atom, such as we discussedback in Chapter 15 with the photoelectric effect. In an oxidation/reduction (collectivelycalled redox) process, if an atom or a molecule is oxidized, then something else isreduced, so that charge is balanced in the end. Likewise, if something is reduced, thensomething else is oxidized.Let’s consider for a moment the electronic structure of Cu, and try to understandthese various oxidation states. If we used the Aufbau principle, Hund’s rule, Pauliprinciple, etc., we would say that the electronic structure of Cuo is 4S2 3D9. Now if weadd one more electron, we can get a closed shell, or 4S2 3D10. This would be Cu-, and wewould expect it to be fairly stable (it is, after all, a closed shell anion). Now imagine ifwe removed one electron from Cuo. We would then have 4S2 3D8. This arrangement ofelectrons doesn’t look especially remarkable – and so one might not expect it to be verystable. However, try moving both of the 4S electrons into the D shell, and you get 4S03D10. This is a closed shell cation, and the electronic configuration now does look ratherspecial! 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 4S23D7 or of 4S0 3D9, depending on how we think about the +2 oxidation state. Note thathere I have been rather casual about moving the 4S electrons back and forth between the3D orbitals. In fact, this is a fine thing to do, and it is very common for transition metalsto 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 twoelectronic configurations would be particularly stable. In fact, Cu++ does exist, but it isnot as common as the other oxidation states. From our simple arguments, we wouldpredict 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 thebeauty and frustration of inorganic chemistry. It would be great if we could just take theelectronic 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. Thefact is that we can’t do this. However, we can use the electronic configurations as aguide to which oxidation states will be most likely. However, lots of possibilities may beobserved – possibilities that we don’t think of when we just look for ‘magic’ electronicconfigurations.What do we mean by ‘magic’ electronic configurations? Here are a few, withaccompanying explanations. The ones highlighted in red are especially stable.4S0 3D0rare gas configuration (ex: Sc+2, Ti+4, Mn+7)4S2 3D0closed S-shell (ex: Ti(II), V(III), Mn(V))4S0 3D5½ filled D-shell (ex: Fe+3, Mn+2)4S1 3D56 unpaired, aligned electrons (example Cro, V-, Fe+2)4S2 3D5closed S-shell, aligned and unpaired electrons in D shell(ex: Co+2, Ni+3)4S0 3D10closed D-Shell (Cuo, Ago, Auo)4S2 3D10closed 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 elementsusing Roman numerals, such as Mn(V). This is equivalent to Mn+5, and is actually amore common way of representing the +5 oxidation state of Manganese. We will useRoman numerals throughout this discussion to represent positive oxidation states oftransition metals.Note also that in some instances, the system that represents the ‘magic’configuration is a neutral atom, such as Cro for 4S1 3D5. In fact, in those cases, the baremetal is very stable. Recall that Cr provides a shiny metallic plating on many automotiveparts, or that mercury (Hgo 6S2 5D10) is a stable metal that is used in thermometers andelsewhere.Note also, that there are many examples of ‘magic’ configurations that are notfound. You can figure this out by comparing the above list with Figure 18.4 (p. 663) ofthe text. For 4S1 3D5 one might expect Mn+1. In fact, Mn+1 is not listed on 18.4, and so itis probably not very stable. You ask “Why not?” I am not really sure, but I believe thatit is because the Mn(II) state is so stable that any Mn(I) that you might form isimmediately oxidized to Mn(II)