The Chemistry of d- and f- Block: Thermodynamics Consequences of CFSE
In this section, we will build up again the concept of CFSE as we will look the consequences over CFSE into some observable values, such as hydration enthalpy of M2+ and ionic radius of M2+ ions. Besides that, we can also determine from the value of Δo from hydration enthalpy. In this section also, we will look how CFSE could be used to predict the geometry of coordinated complex to be a tetrahedral or a octahedral complex.
In previous sections, we already discussed about how CFSE can be calculated but until this far we do not really know what this values could tell us. For example in high spin octahedral d1 system, we can calculate the CFSE as 0.4Δo (see Crystal Field Theory) but actually what this value can tell us? What is the evidences that CFSE is exist?
Firstly, we can see the evidence that CFSE is exist from hydration energies of M2+. Hydration energy is defined as enthalpy change for the reaction of one mole of gaseous M2+ ions with an infinite number of water molecules to give an aqueous solution of hydrated metal cations.
The curve of the hydration energies of is shown below.
From the curve above, we see the dotted line which is the predicted values and the values should follow the red line; in fact it does not follow the line. Most complexes are more stable that we would expect where the stability maxima lie on d3 and d8; d0, d5, and d10 do lie on the line. In aquo complexes, it will form high spin as water is a weak field ligand. CFSE rises to maximum at d3 and d8 and CFSE is 0 at d0, d5, and d10. Therefore, it gives a double-humped curve appearance. Furthermore, Sc cannot form Sc2+ so it does not appear on the curve.
Furthermore, we can also find the value of from the curve of hydration energies. The first thing we need to remember that CFSE is the extra stabilisation energy of a complex. Let take example of Ti2+. From the curve, we can find that the experimental value is -2729 kJ mol-1 and the expected value is approximately -2580 kJ mol-1, hence the difference is 149 kJ mol-1. From splitting diagram for Ti2+, we can find that CFSE is 0.8Δo and this value corresponds to the difference between the experimental and the expected value, so Δo is 186 kJ mol-1.
Besides that, the double-humped curve is also appeared in ionic radius of M2+ in MCl2 and in this curve Sc does not appear for the same reason in the hydration energy curve.
The geometry of the complex, either octahedral or tetrahedral, could be predicted from CFSE when we do not really the know formula (if we know, it would be easy). From CFSE, it tells us that octahedral is more favourable than tetrahedral because forming six bonds is more favourable than forming four bonds and CFSE is usually greater for octahedral than tetrahedral. However, we can estimate how much greater the CFSE is. When we try to compare the value of Δo and Δtet by using the equation of Δtet = 4Δo/9, we will the graph as shown below and the CFSEs is in the unit of Δo.
From the graph, it is shown that CFSE usually favours octahedral over tetrahedral, but the degree of favourability depends on the electronic configuration and for d0, d5, and d10 there is no CFSE difference. Hence the order is:
In table below is some example of tetrahedral complexes.
Another case of geometry also appears in first row of [ML4]n+ which can form tetrahedral or square planar. The tendency to go square planar increases as Δo increases: [NiCl4]2- is tetrahedral, [Ni(CN)4]2- is square planar which avoids filling the high energy dx2-y2 orbital. Furthermore, for the same reason Pd and Pt complexes will be square planar (e.g. [PdCl4]2- and [PtCl4]2-).
In previous sections, we already discussed about how CFSE can be calculated but until this far we do not really know what this values could tell us. For example in high spin octahedral d1 system, we can calculate the CFSE as 0.4Δo (see Crystal Field Theory) but actually what this value can tell us? What is the evidences that CFSE is exist?
Firstly, we can see the evidence that CFSE is exist from hydration energies of M2+. Hydration energy is defined as enthalpy change for the reaction of one mole of gaseous M2+ ions with an infinite number of water molecules to give an aqueous solution of hydrated metal cations.
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| The hydration enthalpies of M2+ curve |
Furthermore, we can also find the value of from the curve of hydration energies. The first thing we need to remember that CFSE is the extra stabilisation energy of a complex. Let take example of Ti2+. From the curve, we can find that the experimental value is -2729 kJ mol-1 and the expected value is approximately -2580 kJ mol-1, hence the difference is 149 kJ mol-1. From splitting diagram for Ti2+, we can find that CFSE is 0.8Δo and this value corresponds to the difference between the experimental and the expected value, so Δo is 186 kJ mol-1.
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| The ionic radius of M2+ in MCl2 |
The geometry of the complex, either octahedral or tetrahedral, could be predicted from CFSE when we do not really the know formula (if we know, it would be easy). From CFSE, it tells us that octahedral is more favourable than tetrahedral because forming six bonds is more favourable than forming four bonds and CFSE is usually greater for octahedral than tetrahedral. However, we can estimate how much greater the CFSE is. When we try to compare the value of Δo and Δtet by using the equation of Δtet = 4Δo/9, we will the graph as shown below and the CFSEs is in the unit of Δo.
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| The curve between Δo and Δtet |
d0, d5, d10 < d1, d6 < d2, d7 < d4, d9 < d3, d8.The units are Δo, so things which make Δo bigger will also disfavour tetrahedral complexes, e.g. strong field ligands, 2nd or 3rd row metals, and high oxidation states. Therefore, we can deduce that to make a tetrahedral complex we need to use:
- large, negatively charged, weak field ligands;
- first row metals (unless d0, d5 high spin, d10)
- a configuration where the CFSE difference is small.
In table below is some example of tetrahedral complexes.
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