Amongst the following the most stable complex is:

A) $\ce{[Fe(H2O)6]^3+}$
B) $\ce{[Fe(NH3)6]^3+}$
C) $\ce{[Fe(C2O4)3]^3-}$
D) $\ce{[FeCl6]^3-}$

I know that the answer should be $\ce{[Fe(C2O4)3]^3-},$ but I could not find a satisfactory explanation.

  • 2
    $\begingroup$ It is better to hold something with two hands than with one. Therefore bidentate ligands generally have an advantage over monodentate. Also, welcome to Chem.SE. $\endgroup$ Aug 23, 2016 at 11:10

1 Answer 1


This is known as the chelate effect. Complexes with bi- or polydentate ligands — chelating ligands to be exact — are observed to be more stable than their nonchelating counterparts.

To understand this, we need to compare apples with apples, so your example is not the best one. (However, based on the trend we can still give a prediction; more on that later.) Instead, I will base my argument on comparing $\ce{[Fe(NH3)6]^3+}$ and $\ce{[Fe(en)3]^3+}$, where $\ce{en}$ is ethylenediamine $\ce{H2N-CH2-CH2-NH2}$.

Consider the following equilibrium:

$$\ce{[Fe(NH3)6]^3+ + 3 en <=> [Fe(en)3]^3+ + 6 NH3}\tag{1}$$

To determine which side is favoured (‘more stable’), we need to take a look at the equilibrium constant $K$. This is defined as

$$K = \frac{[\ce{[Fe(en)3]^3+}][\ce{NH3}]^6}{[\ce{[Fe(NH3)6]^3+}][\ce{en}]^3}\tag{2}$$

But to get a hint on its numeric value without measuring it, we need to apply the Gibbs-Helmholtz equation:

$$\Delta_\mathrm{r}G^⦵ = - RT \ln K\tag{3}$$

Equation (3) tells us that $\Delta_\mathrm{r}G^⦵$ will be negative (exergonic) if $\ln K$ is positive ($R$ and $T$ are positive by definition) — so if $K >1$, i.e. the product side is more favoured than the reactant side. So far no new information. We need to apply another equation, the definition of the Gibbs free energy:

$$\Delta_\mathrm{r}G^⦵ = \Delta_\mathrm{r}H^⦵ - T\Delta_\mathrm{r}S^⦵ \tag{4}$$

For the enthalpic term $\Delta_\mathrm{r}H^⦵$, we need to ask ourselves which bonds are broken and which are reformed. We break six $\ce{Fe\bond{<-}N}$ bonds and form six $\ce{Fe\bond{<-}N}$ bonds over the course of the reaction. Thus we can assume:

$$\Delta_\mathrm{r}H^⦵ \approx 0\tag{5}$$


$$\Delta_\mathrm{r}G^⦵ = - T\Delta_\mathrm{r}S^⦵ \tag{4'}$$

The entropic term $\Delta_\mathrm{r}S^⦵$ is most influenced by the number of particles in solution. The reactant side has four free particles in solution while the product side has seven. Since $7>4$, we can assume the product side’s entropy to be larger. Thus giving:

$$\begin{align}\Delta_\mathrm{r}S^⦵ > 0\tag{6}\\ - T \Delta_\mathrm{r}S^⦵ < 0\tag{7}\\ \Delta_\mathrm{r}G^⦵ < 0\tag{4''}\end{align}$$

Therefore, the reaction as written in equation (1) is exergonic, the products are more stable then the reactants.

Applying this to the problem, we see that most of the ligands are similar enough to not warrant any other stronger effects. Thus, the complex with three bidentate, chelating ligands is most stable.


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