5.3.1 (h,i) Ligand Substitution and Haemoglobin
Syllabus
(h) ligand substitution reactions and the accompanying colour changes in the formation of:
(i) [Cu(NH3)4(H2O)2]2+ and [CuCl4] 2– from [Cu(H2O)6] 2+
(ii) [Cr(NH3)6] 3+ from [Cr(H2O)6] 3+
{Complexed formulae should be used in ligand substitution equations.}
(i) explanation of the biochemical importance of iron in haemoglobin, including ligand substitution involving O2 and CO
What does this mean?
All dative bonds are of a similar strength and so it is relatively easy to substitute one ligand for another, often by simply placing a complex in a concentrated solution of a alternative ligands.
Copper (II) and Ammonia
Adding a a small amount of low concentration ammonia to Copper ions causes precipitation of a light blue solid, more concentrated ammonia will result in the precipitate re-dissolving and being replaced by a dark blue solution.
Why?
Aqueous Copper (II) ions exist as hexaqua complexes [Cu(H2O)6]2+ - they retain the 2+ charge because the ligands are all neutral.
It's this charge that allows the complex to interact with the polar water molecules (that it's not already bonded directly to) and to dissolve.
Adding a base removes H+ (acidic!) from the water ligands until they lose all charge.
[Cu(H2O)6]2+(aq) + NH3(aq)--> [Cu(H2O)5OH]+(aq) + NH4+(aq)
[Cu(H2O)5OH]+(aq) + NH3(aq) --> [Cu(H2O)4(OH)2](s) + NH4+(aq)
Overall reaction = [Cu(H2O)6]2+(aq) + 2 NH3(aq)--> [Cu(H2O)4(OH)2](s) + 2 NH4+(aq)
The complex precipitates out of the solution when it has no more charge - the two OH- ions cancel the charge of the central Cu 2+ metal ion - you could do the same by adding NaOH
However, adding greater and greater concentrations of Ammonia pushes the equilibrium to lower the concentration of Ammonia (Le Chatalier!) by exchanging some of the ligands on the complex.
Note, examiners will accept violet as a colour for this complex but dark blue is what you will likely observe when doing this experiment.
Quite why it finishes doing this when it reaches [Cu(NH3)4(H2O)2]2+ is beyond A level, but you should note that the charge is again 2+ because all the ligands are neutral again.
The colour is somewhat different despite not changing the oxidation state of the metal because it now has different ligands.
Copper (II) and Chloride ions
Adding a concentrated solution of Chloride ions (from conc. HCl or NaCl usually) causes total ligand replacement - all the water ligands are replaced (although not by so many chloride ligands as these can't fit around the Copper ion).
[Cu(H2O)6]2+(aq) + 4Cl-(aq) --> [CuCl4]2-(aq)
Note that the colour changes from blue to Yellow, although you may see green if there is still a high concentration of blue [Cu(H2O)6]2+(aq) solution muixed with the [CuCl4]2-(aq).
The new complex has a 2- charge as it has four 1- ligands and a 2+ central ion.
The colour has changed because the ligands have changed and the coordination number of the complex.
Chromium (III) and Ammonia
Adding a a small amount of low concentration ammonia to Chromium ions causes precipitation too, more concentrated ammonia will result in the precipitate re-dissolving and being replaced by a purple solution.
Why?
Aqueous Chromium (III) ions exist as hexaqua complexes [Cr(H2O)6]3+ - they retain the 3+ charge because the ligands are all neutral.
It's this charge that allows the complex to interact with the polar water molecules (that it's not already bonded directly to) and to dissolve.
Adding a base removes H+ (acidic!) from the water ligands until they lose all charge.
[Cr(H2O)6]3+(aq) + NH3(aq)--> [Cr(H2O)5OH]2+(aq) + NH4+(aq)
[Cr(H2O)5OH]2+(aq) + NH3(aq) --> [Cr(H2O)4(OH)2]+(aq) + NH4+(aq)
[Cr(H2O)4(OH)2]+(aq) + NH3(aq) --> [Cr(H2O)3(OH)3](s) + NH4+(aq)
Overall reaction = [Cr(H2O)6]3+(aq) + 3 NH3(aq)--> [Cr(H2O)3(OH)3](s) + 3 NH4+(aq)
The complex precipitates out of the solution when it has no more charge - the three OH- ions cancel the charge of the central Cr3+ metal ion - you could do the same by adding NaOH
However, adding greater and greater concentrations of Ammonia pushes the equilibrium to lower the concentration of Ammonia (Le Chatalier!) by exchanging some of the ligands on the complex.
Haemoglobin
Haemoglobin is an Iron (II) complex.
The globin carries metal ions through a dative bond from one Nitrogen atom.
Four other Nitrogen atoms do the same from a porphrin ring, leaving one space open to carry a ligand.
This can be water, but can equally well be Oxygen which is easy to pick up in Oxygen-rich lungs.
But equally easily lost in Oxygen poor areas of the body - hence it can deliver Oxygen to where it is needed.
Unless, of course, another ligand occupies the active site and is not easily removed.
Cyanide would have this effect but so would Carbon Monoxide.
Clearly, if enough red blood cells are affected then there are too few to carry oxygen around the body and we would suffocate.
So Carbon Monoxide is a threat to life and is easily produced in poorly ventilated gas-fires etc - it is also colourless and odourless, making a detector a wise investment.