THE GENERAL FEATURES OF TRANSITION METAL CHEMISTRY General features of transition metals вЂў вЂў вЂў вЂў This presentation explains what a transition metal is in terms of its electronic structure, and then goes on to look at the general features of transition metal chemistry. These include variable oxidation state (oxidation number) complex ion formation coloured ions catalytic activity What is a transition metal? вЂў The terms transition metal (or element) and d block element are sometimes used as if they mean the same thing. вЂў They don't - there's a subtle difference between the two terms. d block elements вЂў You will remember that when you are building the Periodic Table and working out where to put the electrons, something odd happens after argon. вЂў At argon, the 3s and 3p levels are full, but rather than fill up the 3d levels next, the 4s level fills instead to give potassium and then calcium. вЂў Only after that do the 3d levels fill. d block elements The elements in the Periodic Table which correspond to the d sublevels filling are called d block elements. The first row of these is shown in the shortened form of the Periodic Table below. The electronic structures of the d block elements shown are: Sc [Ar] 3d14s2 Ti [Ar] 3d24s2 V [Ar] 3d34s2 Cr [Ar] 3d54s1 Mn [Ar] 3d54s2 Fe [Ar] 3d64s2 Co [Ar] 3d74s2 Ni [Ar] 3d84s2 Cu [Ar] 3d104s1 Zn [Ar] 3d104s2 You will notice that the pattern of filling isn't entirely tidy! It is broken at both chromium and copper. This is something that you are just going to have to accept. There is no simple explanation for it which is usable at this level. Any simple explanation which is given is faulty! People sometimes say that a half-filled d level as in chromium (with one electron in each orbital) is stable, and so it is - sometimes! But you then have to look at why it is stable. The obvious explanation is that chromium takes up this structure because separating the electrons minimises the repulsions between them - otherwise it would take up some quite different structure. But you only have to look at the electronic configuration of tungsten (W) to see that this apparently simple explanation doesn't always work. Tungsten has the same number of outer electrons as chromium, but its outer structure is different 5d46s2. Again the electron repulsions must be minimised otherwise it wouldn't take up this configuration. But in this case, it isn't true that the half-filled state is the most stable - it doesn't seem very reasonable, but it's a fact! The real explanation is going to be much more difficult than it seems at first sight. Neither can you use the statement that a full d level (for example, in the copper case) is stable, unless you can come up with a proper explanation of why that is. You can't assume that looking nice and tidy is a good enough reason! If you can't explain something properly, it is much better just to accept it than to make up faulty explanations which sound OK on the surface but don't stand up to scrutiny! Transition metals вЂў Not all d block elements count as transition metals! There are discrepancies between the various syllabuses, but the majority use the definition: вЂў A transition metal is one which forms one or more stable ions which have incompletely filled d orbitals. SCANDIUM , ZINC and COPPER On the basis of this definition, scandium and zinc don't count as transition metals - even though they are members of the d block. Scandium has the electronic structure [Ar] 3d14s2. When it forms ions, it always loses the 3 outer electrons and ends up with an argon structure. The Sc3+ ion has no d electrons and so doesn't meet the definition. Zinc has the electronic structure [Ar] 3d104s2. When it forms ions, it always loses the two 4s electrons to give a 2+ ion with the electronic structure [Ar] 3d10. The zinc ion has full d orbitals and doesn't meet the definition either. By contrast, copper, [Ar] 3d104s1, forms two ions. In the Cu+ ion the electronic structure is [Ar] 3d10. However, the more common Cu2+ ion has the structure [Ar] 3d9. Copper is definitely a transition metal because the Cu2+ ion has an incomplete d orbitals. Transition metal ions You have already come across the fact that when the Periodic Table is being built, the 4s orbital is filled before the 3d orbitals. This is because before filling orbitals, 4s orbitals have a lower energy than 3d orbitals. However, once the electrons are actually in their orbitals, the energy order changes - and in all the chemistry of the transition elements, the 4s orbital behaves as the outermost, highest energy orbital. Transition metal ions The reversed order of the 3d and 4s orbitals only applies to building the atom up in the first place. In all other respects, you treat the 4s electrons as being the outer electrons. This is another of those things that you just have to accept. The explanation again lies well beyond the level you are working at. Just remember that once you have the full electronic structure for one of these atoms, the 4s electrons are the outermost electrons. Remember this: When d-block elements form ions, the 4s electrons are lost first. Transition metal ions To write the electronic structure for Co2+: Co [Ar] 3d74s2 Co2+ [Ar] 3d7 The 2+ ion is formed by the loss of the two 4s electrons. To write the electronic structure for V3+: V [Ar] 3d34s2 V3+ [Ar] 3d2 The 4s electrons are lost first followed by one of the 3d electrons. Transition metal ions To write the electronic structure for Cr3+: Cr Cr3+ 1s22s22p63s23p63d54s1 1s22s22p63s23p63d3 The 4s electron is lost first followed by two of the 3d electrons. To write the electronic structure for Zn2+: Zn Zn2+ 1s22s22p63s23p63d104s2 1s22s22p63s23p63d10 This time there is no need to use any of the 3d electrons. To write the electronic structure for Fe3+: Fe Fe3+ 1s22s22p63s23p63d64s2 1s22s22p63s23p63d5 Forming transition metal ions The rule is quite simple. Take the 4s electrons off first, and then as many 3d electrons as necessary to produce the correct positive charge. Variable oxidation state (number) One of the key features of transition metal chemistry is the wide range of oxidation states (oxidation numbers) that the metals can show. It would be wrong, though, to give the impression that only transition metals can have variable oxidation states. For example, elements like sulphur or nitrogen or chlorine have a very wide range of oxidation states in their compounds - and these obviously aren't transition metals. However, this variability is less common in metals apart from the transition elements. Of the familiar metals from the main groups of the Periodic Table, only lead and tin show variable oxidation state to any extent. Examples of variable oxidation states in the transition metals Iron Iron has two common oxidation states (+2 and +3) in, for example, Fe2+ and Fe3+. It also has a less common +6 oxidation state in the ferrate(VI) ion, FeO42-. Manganese Manganese has a very wide range of oxidation states in its compounds. For example: +2 in Mn2+ +3 in Mn2O3 +4 in MnO2 +6 in MnO42+7 in MnO4- Explaining the variable oxidation states in the transition metals We'll look at the formation of simple ions like Fe2+ and Fe3+. When a metal forms an ionic compound, the formula of the compound produced depends on the energetics of the process. On the whole, the compound formed is the one in which most energy is released. The more energy released, the more stable the compound. There are several energy terms to think about, but the key ones are: п‚· п‚· The amount of energy needed to ionise the metal (the sum of the various ionisation energies) The amount of energy released when the compound forms. This will either be lattice enthalpy if you are thinking about solids, or the hydration enthalpies of the ions if you are thinking about solutions. The more highly charged the ion, the more electrons you have to remove and the more ionisation energy you will have to provide. But off-setting this, the more highly charged the ion, the more energy is released either as lattice enthalpy or the hydration enthalpy of the metal ion. Thinking about a typical non-transition metal (calcium) Calcium chloride is CaCl2. Why is that? If you tried to make CaCl, (containing a Ca+ ion), the overall process is slightly exothermic. By making a Ca2+ ion instead, you have to supply more ionisation energy, but you get out lots more lattice energy. There is much more attraction between chloride ions and Ca2+ ions than there is if you only have a 1+ ion. The overall process is very exothermic. Because the formation of CaCl2 releases much more energy than making CaCl, then CaCl2 is more stable - and so forms instead. Thinking about a typical nontransition metal (calcium) What about CaCl3? This time you have to remove yet another electron from calcium. The first two come from the 4s level. The third one comes from the 3p. That is much closer to the nucleus and therefore much more difficult to remove. There is a large jump in ionisation energy between the second and third electron removed. Although there will be a gain in lattice enthalpy, it isn't anything like enough to compensate for the extra ionisation energy, and the overall process is very endothermic. It definitely isn't energetically sensible to make CaCl3! Thinking about a typical transition metal (iron) Here are the changes in the electronic structure of iron to make the 2+ or the 3+ ion. Fe [Ar] 3d64s2 Fe2+ [Ar] 3d6 Fe3+ [Ar] 3d5 The 4s orbital and the 3d orbitals have very similar energies. There isn't a huge jump in the amount of energy you need to remove the third electron compared with the first and second. The figures for the first three ionisation energies (in kJ mol-1) for iron compared with those of calcium are: metal 1st IE 2nd IE 3rd IE Ca 590 1150 4940 Fe 762 1560 2960 There is an increase in ionisation energy as you take more electrons off an atom because you have the same number of protons attracting fewer electrons. Thinking about a typical transition metal (iron) However, there is much less increase when you take the third electron from iron than from calcium. In the iron case, the extra ionisation energy is compensated more or less by the extra lattice enthalpy or hydration enthalpy evolved when the 3+ compound is made. The net effect of all this is that the overall enthalpy change isn't vastly different whether you make, say, FeCl2 or FeCl3. That means that it isn't too difficult to convert between the two compounds. The formation of complex ions What is a complex ion? A complex ion has a metal ion at its centre with a number of other molecules or ions surrounding it. These can be considered to be attached to the central ion by co-ordinate (dative covalent) bonds. (In some cases, the bonding is actually more complicated than that.) The molecules or ions surrounding the central metal ion are called ligands. Simple ligands include water, ammonia and chloride ions. What all these have got in common is active lone pairs of electrons in the outer energy level. These are used to form co-ordinate bonds with the metal ion. Some examples of complex ions formed by transition metals [Fe(H2O)6]2+ [Co(NH3)6]2+ [Cr(OH)6]3[CuCl4]2Other metals also form complex ions - it isn't something that only transition metals do. Transition metals do, however, form a very wide range of complex ions. Bonding in simple complex ions Al(H2O)6 3+ We are going to look in detail at the bonding in the complex ion formed when water molecules attach themselves to an aluminium ion to give Al(H2O)63+. Start by thinking about the structure of a naked aluminium ion before the water molecules bond to it. Aluminium has the electronic structure 1s22s22p63s23px1 When it forms an Al3+ ion it loses the 3-level electrons to leave 1s22s22p6 Bonding in simple complex ions That means that all the 3-level orbitals are now empty. The aluminium uses of six of these to accept lone pairs from six water molecules. It re-organises (hybridises) the 3s, the three 3p, and two of the 3d orbitals to produce six new orbitals all with the same energy. You might wonder why it chooses to use six orbitals rather than four or eight or whatever. Six is the maximum number of water molecules it is possible to fit around an aluminium ion (and most other metal ions). By making the maximum number of bonds, it releases most energy and so becomes most energetically stable. Only one lone pair is shown on each water molecule. The other lone pair is pointing away from the aluminium and so isn't involved in the bonding. The resulting ion looks like this: Dotted arrows represent lone pairs coming from water molecules behind the plane of the screen or paper. Wedge shaped arrows represent bonds from water molecules in front of the plane of the screen or paper. Al(H2O)6 3+ Because of the movement of electrons towards the centre of the ion, the 3+ charge is no longer located entirely on the aluminium, but is now spread over the whole of the ion. Because the aluminium is forming 6 bonds, the coordination number of the aluminium is said to be 6. The co-ordination number of a complex ion counts the number of co-ordinate bonds being formed by the metal ion at its centre. In a simple case like this, that obviously also counts the number of ligands - but that isn't necessarily so, as you will see later. Some ligands can form more than one coordinate bond with the metal ion. Fe(H2O)6 3+ This example is chosen because it is very similar to the last one - except that it involves a transition metal. Iron has the electronic structure 1s22s22p63s23p63d64s2 When it forms an Fe3+ ion it loses the 4s electrons and one of the 3d electrons to leave 1s22s22p63s23p63d5 Looking at this as electrons-in-boxes, at the bonding level: Fe(H2O)6 +3 Looking at this as electrons-in-boxes, at the bonding level: Now, be careful! The single electrons in the 3d level are NOT involved in the bonding in any way. Instead, the ion uses 6 orbitals from the 4s, 4p and 4d levels to accept lone pairs from the water molecules. Before they are used, the orbitals are re-organised (hybridised) to produce 6 orbitals of equal energy. Once the co-ordinate bonds have been formed, the ion looks exactly the same as the equivalent aluminium ion. Because the iron is forming 6 bonds, the co-ordination number of the iron is 6. CuCl4 2- This is a simple example of the formation of a complex ion with a negative charge. Copper has the electronic structure 1s22s22p63s23p63d104s1 When it forms a Cu2+ ion it loses the 4s electron and one of the 3d electrons to leave 1s22s22p63s23p63d9 To bond the four chloride ions as ligands, the empty 4s and 4p orbitals are used (in a hybridised form) to accept a lone pair of electrons from each chloride ion. Because chloride ions are bigger than water molecules, you can't fit 6 of them around the central ion - that's why you only use 4. Only one of the 4 lone pairs on each chloride ion is shown. The other three are pointing away from the copper ion, and aren't involved in the bonding. CuCl42That gives you the complex ion: The ion carries 2 negative charges overall. That comes from a combination of the 2 positive charges on the copper ion and the 4 negative charges from the 4 chloride ions. In this case, the co-ordination number of the copper is, of course, 4.