Lecture Nineteen - Coordination complexes continued Flashcards
What do all first row transition metals do?
All first row transition metal ions dissolve in water to form complexes with aqua or hydroxy ligands.
[M(H2O)6]2+ complexes of all the first row d-block metal (II) ions are known.
Discuss lewis acids and charge density with regards to coordination chemistry.
Some metal cations are strong Lewis acids, e.g. Li+, Mg2+, Al3+, Fe3+ and Ti3+.
Charge density of an ion = Charge on the ion / Surface area of the ion.
Acidity of aqueous solution containing the metal cation depends on the position of the equilibrium.
E.g. [V(H2)6]2+ is rapidly oxidised in air which leads to two [V(H2O)5(HO)]2+ complexes combine and eliminate water.
This new complex has changed nuclearity and is a dinuclear cation.
Also the two hydroxy ligans are now bridging, hence the greek symbol mu is used.
I.e. New molecule formed is [(H2O)4V(μ-OH)2V(H2O)4]4+.
Explain complex ammines in coordination complexes.
Addition of aqueous ammonia to most [M(H2O)6]2+ causes displacement of the water ligands.
These ligand displacement reactions are equilibrium reactions.
Explain chloridos in coordination complexes.
Reactions of metal salts with excess chloride (Cl-) generally give tetrahedral chlorido complexes.
Explain cyanido coordination complexes.
The CN- ligand usually binds via the carbon atom, as M-C(triple bond) N, which is linear, although it an bridge with both atoms binding.
Fe(II) formes a stable [Fe(CN)6]4- salt, which when added to a solution of Fe3+ forms a deef blue ‘Prussian Blue’ a qualitative test.
What are bidentate ligands?
Bidentate ligands can be symmetrical or asymmetrical.
Three symmetrical bidentate ligands forming an octahedral gives enantiomers.
And if the precursor is achiral the product will be racemic, i.e. equal amounts of the two enantiomers.
What are polydentate ligands?
The polydentate ligans, H4edta, wraps around the metal ion to form an octahedral complex.
Explain coordination omplex stability.
When do ligands exchange and why?
In chemical equilibrium we can meaure the equilibrium constant when each ligand is exchaned for another in solution.
So, for two complex ions we can compare their equilibrium constant, K or stability constant.
E.g. The replacement of water by ammonia happens in a stepwise manner.
[Cu(H2O)6]2+(aq) + 4NH3(aq) <–>[Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l)
Therefore there is one equilibrium constnat for each water replaced by ammonia.
The overall stability constant can be described as:
- The formaion constant Kf.
- Or can be distingiushed by the use of βy , where y is the number of exchane steps.
Gibbs free energy can be found by using the equation:
∆G° = -RT ln K
