This section 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 that include the following:

· Variable oxidation state (oxidation number)
· Complex ion formation
· Coloured ions
· Catalytic activity

The electronic structures of transition metals

What is a transition metal?
The terms transition metal (or element) and d block element are sometimes used as if they mean the same thing. They don't - there's a subtle difference between the two terms. We'll explore d block elements first:

d block elements

You will remember that when you are building the Periodic Table and working out where to put the electrons, something odd happens after argon.

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. Only after that do the 3d levels fill.

The elements in the Periodic Table which correspond to the d levels 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:



You will notice that the pattern of filling isn't entirely tidy! It is broken at both chromium and copper.

Note: Why is this so? 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.

Transition metals
Not all d block elements count as transition metals! There are discrepancies between various syllabuses, but the majority use the definition:

A transition metal is one which forms one or more stable ions which have incompletely filled d orbitals.

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 levels 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 level.

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 in the empty atom, 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.

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.

Remember this:

When d-block elements form ions, the 4s electrons are lost first.

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.

In this seciton, we will be contrasting the properties of transition elements with calcium.

Properties covered:
- melting point
- density
- atomic radius
- ionic radius
- first ionisation energy
- conductivity with calcium

Melting point of transition metals VS Calcium

Melting point of transition metals tends to be higher than calcium and other s-block metals due to the presence of delocalized electrons in the d and s orbitals. The greater number of delocalized electrons in transition elements as compared to calcium leads to stronger metallic bonds being formed. For example the valence shell of calcium ion has 4s2 configuration. However, the 3d54s1 electronic configuration for Chromium ( Cr) in the first series of transition metals is more stable and so produces stronger bonds than the 4s2 electronic configuration in calcium.

Therefore, more heat energy is required to break the metallic bonds in transition metals than calcium leading to higher melting points.

Atomic radii and density of transition metals VS Calcium
Densities of transition metals are much greater than densities of s-block metals like calcium. There is a gradual increase in the densities of transition metals from scandium (2.99g/cm^3) to copper (8.92g/cm^3). The greater densities of transition metals as compared to typical s-block metals s are due to their smaller atomic radii coupled with the greater atomic masses. For example, the density of calcium is 1.6g/cm^3.

As density is calculated by dividing the atomic mass of the metals by their atomic radii, transition metals consequently have higher densities than the s-block metals like calcium.

Ionic radius

First ionisation energy

The first ionisation energy is the energy required to remove the most loosely held electron from one mole of gaseous atoms to produce 1 mole of gaseous ions each with a charge of 1+.All elements have a first ionisation energy

Conductivity of transition metals VS calcium

The delocalized electrons in a metal acts as as mobile charge carriers. Transition metals have electrons in at least the d orbital. While calcium only has electrons until the S orbital because of its electronic configuration(1s2 2s2 2p6 3s2 3p6 4s2 ).

As transition metals have more delocalized electons( mobile charge carriers) than calcium, they have a higher electrical conductivity than calcium.

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:

· 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 ation 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.

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:





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. 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!


Common Oxidation States of the First Series of Transition Metals

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]2-

Other 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.

Why do we see some compounds as being coloured?

White light

You will know, of course, that if you pass white light through a prism it splits into all the colours of the rainbow. Visible light is simply a small part of an electromagnetic spectrum most of which we can't see - gamma rays, X-rays, infra-red, radio waves and so on.

Each of these has a particular wavelength, ranging from 10-16 metres for gamma rays to several hundred metres for radio waves. Visible light has wavelengths from about 400 to 750 nm. (1 nanometre = 10-9 metres.)

The diagram shows an approximation to the spectrum of visible light.





Why is copper(II) sulphate solution blue?

If white light (ordinary sunlight, for example) passes through copper(II) sulphate solution, some wavelengths in the light are absorbed by the solution. Copper(II) ions in solution absorb light in the red region of the spectrum.

The light which passes through the solution and out the other side will have all the colours in it except for the red. We see this mixture of wavelengths as pale blue (cyan).

The diagram gives an impression of what happens if you pass white light through copper(II) sulphate solution.






Working out what colour you will see isn't easy if you try to do it by imagining "mixing up" the remaining colours. You wouldn't have thought that all the other colours apart from some red would look cyan, for example.


Sometimes what you actually see is quite unexpected. Mixing different wavelengths of light doesn't give you the same result as mixing paints or other pigments.
You can, however, sometimes get some estimate of the colour you would see using the idea of complementary colours.


Complementary colours


If you arrange some colours in a circle, you get a "colour wheel". The diagram shows one possible version of this. An internet search will throw up many different versions!







Colours directly opposite each other on the colour wheel are said to be complementary colours. Blue and yellow are complementary colours; red and cyan are complementary; and so are green and magenta.


Mixing together two complementary colours of light will give you white light.


Beware: That is NOT the same as mixing together paint colours. If you mix yellow and blue paint you don't get white paint. (:


What this all means is that if a particular colour is absorbed from white light, what your eye detects by mixing up all the other wavelengths of light is its complementary colour. Copper (II) sulphate solution is pale blue (cyan) because it absorbs light in the red region of the spectrum. Cyan is the complementary colour of red.


Some sample colours
The diagrams show the approximate colours of some typical metal ions. The charge on these ions is typically 2+ or 3+.


Non-transition metal ions





These ions are all colourless. (Sorry, we can't do genuinely colourless!)

Transition metal ions




The corresponding transition metal ions are coloured. So . . . what causes transition metal ions to absorb wavelengths from visible light (causing colour) whereas non-transition metal ions don't? And why does the colour vary so much from ion to ion?



The origin of colour in complex ions containing transition metals

Complex ions containing transition metals are usually coloured, whereas the similar ions from non-transition metals aren't. That suggests that the partly filled d orbitals must be involved in generating the colour in some way. Remember that transition metals are defined as having partly filled d orbitals.

Octahedral complexes


For simplicity we are going to look at the octahedral complexes which have six simple ligands arranged around the central metal ion. The argument isn't really any different if you have multidentate ligands - it's just slightly more difficult to imagine!


When the ligands bond with the transition metal ion, there is repulsion between the electrons in the ligands and the electrons in the d orbitals of the metal ion. That raises the energy of the d orbitals.


However, because of the way the d orbitals are arranged in space, it doesn't raise all their energies by the same amount. Instead, it splits them into two groups.


The diagram shows the arrangement of the d electrons in a Cu2+ ion before and after six water molecules bond with it.








Whenever 6 ligands are arranged around a transition metal ion, the d orbitals are always split into 2 groups in this way - 2 with a higher energy than the other 3.


The size of the energy gap between them (shown by the blue arrows on the diagram) varies with the nature of the transition metal ion, its oxidation state (whether it is 3+ or 2+, for example), and the nature of the ligands.

When white light is passed through a solution of this ion, some of the energy in the light is used to promote an electron from the lower set of orbitals into a space in the upper set.

Each wavelength of light has a particular energy associated with it. Red light has the lowest energy in the visible region. Violet light has the greatest energy.
Suppose that the energy gap in the d orbitals of the complex ion corresponded to the energy of yellow light.




Each wavelength of light has a particular energy associated with it. Red light has the lowest energy in the visible region. Violet light has the greatest energy.



Suppose that the energy gap in the d orbitals of the complex ion corresponded to the energy of yellow light.








The yellow light would be absorbed because its energy would be used in promoting the electron. That leaves the other colours.

Your eye would see the light passing through as a dark blue, because blue is the complementary colour of yellow.





What about non-transition metal complex ions?

Non-transition metals don't have partly filled d orbitals. Visible light is only absorbed if some energy from the light is used to promote an electron over exactly the right energy gap. Non-transition metals don't have any electron transitions which can absorb wavelengths from visible light.


For example, although scandium is a member of the d block, its ion (Sc3+) hasn't got any d electrons left to move around. This is no different from an ion based on Mg2+ or Al3+. Scandium(III) complexes are colourless because no visible light is absorbed.

In the zinc case, the 3d level is completely full - there aren't any gaps to promote an electron in to. Zinc complexes are also colourless.


Tetrahedral complexes

Simple tetrahedral complexes have four ligands arranged around the central metal ion. Again the ligands have an effect on the energy of the d electrons in the metal ion. This time, of course, the ligands are arranged differently in space relative to the shapes of the d orbitals.

The net effect is that when the d orbitals split into two groups, three of them have a greater energy, and the other two a lesser energy (the opposite of the arrangement in an octahedral complex).

Apart from this difference of detail, the explanation for the origin of colour in terms of the absorption of particular wavelengths of light is exactly the same as for octahedral complexes.


Note: Coding for the ligand


The table shows some common ligands and the code for them in the name of a complex ion. The old names sometimes differ by a letter or so, but never enough for it to be confusing.








Take care with the code for ammonia as a ligand - it has 2 "m"s in its name. If you miss one of these out so that you are left with "amine" or "amino", you are referring to the NH2 group in an organic compound.


This is probably the only point of confusion with these names.

The factors affecting the colour of a transition metal complex ion

In each case we are going to choose a particular metal ion for the centre of the complex, and change other factors. Colour changes in a fairly haphazard way from metal to metal across a transition series.





The nature of the ligand

Different ligands have different effects on the energies of the d orbitals of the central ion. Some ligands have strong electrical fields which cause a large energy gap when the d orbitals split into two groups. Others have much weaker fields producing much smaller gaps.



Remember that the size of the gap determines what wavelength of light is going to get absorbed.

The list shows some common ligands. Those at the top produce the smallest splitting; those at the bottom the largest splitting.





The greater the splitting, the more energy is needed to promote an electron from the lower group of orbitals to the higher ones. In terms of the colour of the light absorbed, greater energy corresponds to shorter wavelengths.

That means that as the splitting increases, the light absorbed will tend to shift away from the red end of the spectrum towards orange, yellow and so on.