Chemguide: Support for CIE A level Chemistry


Learning outcome 12.2(c)

This statement is about ligand exchange reactions in transition metal complexes.

Before you go on, you should find and read the statement in your copy of the syllabus.


Ligand exchange reactions

Look at this page about ligand exchange.

You don't need to remember all of this, but you should understand what ligand exchange means. Concentrate particularly on the examples involving copper and cobalt, because those are mentioned on the syllabus in this statement and in statement 12.2(a). But read the rest of the page as well.

Ignore any link which offers to explain anything in more detail!


Complex ions in copper chemistry

These are mentioned in both this syllabus statement and 12.2(a), and come up frequently in exams.

The reaction of hexaaquacopper(II) ions with hydroxide ions

Hydroxide ions (from, say, sodium hydroxide solution) remove hydrogen ions from the water ligands attached to the copper ion. The hydroxide ions are acting as a base, and the hexaaqua ion is acting as an acid. That is typical of ions like this.

Once a hydrogen ion has been removed from two of the water molecules, you are left with a complex with no charge - a neutral complex. This is insoluble in water and a precipitate is formed.


Note:  The colour coding is to show that this isn't a ligand exchange reaction. The oxygens which were originally attached to the copper are still attached in the neutral complex.


This neutral complex is what is normally known as copper(II) hydroxide. We often write this just as Cu(OH)2, ignoring the waters present.

In the test-tube, the colour change is:

Reactions of hexaaquacopper(II) ions with ammonia solution

The ammonia acts as both a base and a ligand. With a small amount of ammonia, hydrogen ions are pulled off the hexaaqua ion exactly as in the hydroxide ion case to give the same neutral complex.

That precipitate dissolves if you add an excess of ammonia.

The ammonia replaces water as a ligand to give tetraamminediaquacopper(II) ions. Notice that only 4 of the 6 water molecules are replaced.


Note:  In fact this reaction is better looked at as the replacement of 4 of the water molecules in the hexaaquacopper(II) ion, [Cu(H2O)6]2+.

However, here I am using the equation for this reaction that is given by the CIE teacher support material and also in the Chemistry Coursebook. In fact, this is a summary of a complicated series of equilibria. If your chemistry is good, and you are interested, you will find the explanation in full on the page about the reactions between hexaaqua ions and ammonia solution. You do not need to know this for exam purposes, though.

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The colour changes are:

It is possible to reverse all of this by adding a dilute acid such as sulphuric acid to the deep blue solution.

The acid first reacts with the ammonia, reversing the last equation. You go back to the neutral complex. The blue precipitate of copper(II) hydroxide is formed again. Adding more acid puts the hydrogen ions back on the two OH groups again, and so you get back to the pale blue soluble hexaaqua complex.


A ligand exchange reaction involving chloride ions

If you add concentrated hydrochloric acid to a solution containing hexaaquacopper(II) ions, the six water molecules are replaced by four chloride ions.

The reaction taking place is reversible.

Because the reaction is reversible, you get a mixture of colours due to both of the complex ions.

You may find the colour of the tetrachlorocuprate(II) ion variously described as olive-green or yellow. If you add water to the green solution, it returns to the blue colour. The equilibrium is easily reversed by adding more water.


Note:   If you dissolve copper(II) oxide in concentrated hydrochloric acid (to make copper(II) chloride which will then go on to form the tetrachlorocuprate(II) ion), you get a dark yellowish brown solution. This probably better reflects the colour of the ion. CIE described it as yellow-green in the only question I could find up to November 2010.


The question that CIE set involving this complex asked about the effect of adding concentrated ammonium chloride solution to copper(II) sulphate solution. You were told that this produced yellow-green crystals with an empirical formula of CuN2H8Cl4, and were asked to explain what was happening.

If you hadn't come across this reaction before, you are likely to get this wrong under the pressure of exam conditions. The crystals formed are actually ammonium tetrachlorocuprate(II), (NH4)2CuCl4.

This is a ligand exchange involving the chloride ions from the concentrated ammonium chloride solution.

The trap to fall into is to think that it must be a ligand exchange involving ammonia - and that's the trap which the Examiner's Report showed that many people did in fact fall into. It looks to me as if the question was deliberately set to persuade people to fall into this trap.

If you stopped to think about it, it can't be an exchange involving ammonia for two reasons. The colour is wrong (the ammino complex is very dark blue), and you have ammonium ions, not ammonia present. Ammonium ions don't have a lone pair of electrons, and so can't act as ligands.

It isn't always easy to think that clearly under exam conditions, so it helps if you have come across this reaction in advance. Well . . . now you have!


Some other bits and pieces of copper chemistry

I include this here, not because it is on the syllabus at this point (it isn't!), but because it has come up in questions about the copper chemistry we have talked about above. It involves the effect of heat on copper(II) carbonate and copper(II) nitrate.

You will have come across the effect of heat on nitrates and carbonates as a part of statement 10.1(f), but this relates to Group 2, and you may not have made the link that it also applies to transition metal carbonates and nitrates.

The effect of heat on copper(II) carbonate

Copper(II) carbonate is a green powder which turns black when heated, and gives off a colourless gas which turns lime water milky (carbon dioxide). The black residue is copper(II) oxide.

The effect of heat on copper(II) nitrate

Hydrated copper(II) nitrate is blue crystals, but the question which asked about this talked about the effect of heat on anhydrous copper(II) nitrate which is also apparently blue. You weren't asked for this colour, and the question didn't give it, either.

Either form turns black when heated, and gives off the brown gas, nitrogen dioxide, and oxygen. Oxygen, of course, relights a glowing splint. The black residue is again copper(II) oxide.


Complex ions in cobalt chemistry

This is very similar to the copper examples above. This is new to the syllabus, and won't be examined until 2016, so it is hard to know what will be asked.

Cobalt complexes are mentioned in the introductory statement 12.2(a), but not in this one, but I would still be inclined to learn the cobalt as well as the copper complexes in ligand exchange reactions.


The reaction of hexaaquacobalt(II) ions with hydroxide ions

As with the copper case, hydroxide ions (from, say, sodium hydroxide solution) remove hydrogen ions from the water ligands attached to the cobalt ion.

Once again, you are left with a neutral complex which is insoluble in water and so a precipitate is formed.

This neutral complex is what is normally known as cobalt(II) hydroxide. We often write this just as Co(OH)2, ignoring the waters present.

In the test-tube, the colour change is:


Note:  I don't know the reason for the change in colour. One reliable source quotes it vaguely as a change in coordination of the metal ion. Another suggests that the blue precipitate involves a chloride ion (from the cobalt(II) chloride solution if that's what you used) as a ligand. Don't worry about it.


Reactions of hexaaquacobalt(II) ions with ammonia solution

The ammonia acts as both a base and a ligand. With a small amount of ammonia, hydrogen ions are pulled off the hexaaqua ion exactly as in the hydroxide ion case to give the same neutral complex.

That precipitate dissolves if you add an excess of ammonia. The ammonia replaces water as a ligand to give hexaamminecobalt(II) ions.

The colour changes are:

You don't need to worry about the change on standing. You do need to realise that in this case all 6 of the water ligands are replaced by ammonia, unlike the copper case.


A ligand exchange reaction involving chloride ions

If you add concentrated hydrochloric acid to a solution containing hexaaquacobalt(II) ions, the solution turns from its original pink colour to a rich blue. The six water molecules are replaced by four chloride ions.

The reaction taking place is reversible.

If you add water to the blue solution, it returns to the pink colour.


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© Jim Clark 2011 (last modified May 2014)