# 10.1.1: Thermodynamic Data

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We can get some insight into the stability of coordination complexes by looking at their formation constants. The formation constant is the equilibrium constant for the reaction leading to the formation of a coordination complex. The greater the formation constant, the more thermodynamically stable the product complex compared to the reactant complex.

Many complexation reactions take place in aqueous solution by dissolving a metal ion in water to produce the "aqueous metal ion". An aqueous metal ion is one that is coordinated by a number of water molecules, and since it is charged, there must be counter ions to balance that charge. When an aqueous metal ion reacts to form a new complex, its formation constant indicates the stability of the product compared to the aqueous complex. Depending on the number of ligands that are replaced in the product compared to the reactant, the formation complex is defined as either a stepwise formation constant ($$K$$) or an overall formation constant ($$\beta$$).

For example, the formation of [Cu(NH3)(OH2)5]2+ complex ion from Cu2+(aq) can be represented by the following stepwise reaction:1

$\ce{[Cu(OH2)6]^2+ + NH3⇌[Cu(NH3)(OH2)5]^2+ + H2O} \qquad \qquad K_1=1.9 \times 10^4\nonumber$

The formation constant, $$K_1$$, is a stepwise formation constant because it represents the substitution of a single aquo ligand (a coordinated water) by an ammine ligand. Another example is the reaction below that represents the substitution of four aquo ligands for four ammine ligands. The overall formation of the $$\ce{[Cu(NH3)4(OH2)2]^2+}$$ complex ion from Cu2+(aq) would be represented by this equation:

$\ce{[Cu(OH2)6]^2+ + 4NH3⇌[Cu(NH3)4(OH2)2]^2+ + 4H2O} \qquad \qquad \beta_4=1.1 \times 10^{13}\nonumber$

This time, the formation constant represents the substitution of four ammine ligands for four aquo ligands. This transformation does not happen in one step, so the formation constant, $$\beta_4$$ is a composite of the equilibrium constants for four individual steps:

$\begin{array}{cl} \ce{[Cu(OH2)6]^2+ + NH3⇌[Cu(NH3)(OH2)5]^2+ + H2O} & K_1=\frac{\ce{[Cu(NH3)(OH2)5^2+]}}{\ce{[Cu(OH2)6^2+][NH3]}}=1.9 \times 10^4 \\ \ce{[Cu(NH3)(OH2)5]^2+ + NH3⇌[Cu(NH3)2(OH2)4]^2+ + H2O} & K_2=\frac{\ce{[Cu(NH3)2(OH2)4^2+]}}{\ce{[Cu(NH3)(OH2)5^2+][NH3]}} =3.9 \times 10^3 \\ \ce{[Cu(NH3)2(OH2)4]^2+ + NH3⇌[Cu(NH3)3(OH2)3]^2+ + H2O} & K_3=\frac{\ce{[Cu(NH3)3(OH2)3^2+]}}{\ce{[Cu(NH3)2(OH2)4^2+][NH3]}}=1.0 \times 10^3 \\ \ce{[Cu(NH3)3(OH2)3]^2+ + NH3⇌[Cu(NH3)4(OH2)2]^2+ + H2O} & K_4=\frac{\ce{[Cu(NH3)4(OH2)2^2+]}}{\ce{[Cu(NH3)3(OH2)3^2+][NH3]}} =1.5 \times 10^2\\ \end{array}\nonumber$

$\beta_4 = K_1\times K_2\times K_3 \times K_4 =(1.9\times 10^4)(3.9\times 10^3)(1.0 \times 10^3)(1.5 \times 10^2)=1.1\times 10^{13}\nonumber$

##### (\beta\) notation

The $$\beta$$ notation is used to distinguish an overall formation constant from a stepwise formation constant. However, sometimes alternative notation will be used. You may see other texts, the literature, or your instructor using alternative ways to distinguish an overall formation constant like $$\beta_4$$ from the stepwise formation constant, like $$K_4$$. The important thing is to understand the conventions used for whatever source you are reading.

In both the examples above, the equilibrium lies to the right (in other words, the formation constant is >> 1). Another way to interpret this is that Cu(II) has a higher affinity for ammonia than for water, and thus the ammine complex is more stable than the aqueous metal ion. In the case of the copper tetraammine complex, replacement of each individual water by ammonia is favorable, so the aggregate formation constant $$\beta_4$$ becomes very large. The preference of Cu(II) for ammonia can be explained by the increased basicity of ammonia compared to water.

Formation constants can provide valuable insight into factors that contribute to the stability of coordination complexes, but we may need a more extensive set of data to build a more complete picture. If we look at formation constants in a series of complexes that have something in common, we may get an idea about additional factors that play a role in complex stability. Take a look at these examples:2

$\begin{array}{cl} \ce{Cu+(aq) +2 Cl- ⇌ [CuCl2]-} & K = 3.0 \times 10^5 \\ \ce{Cu+(aq) + 2 Br- ⇌[CuBr2]-} & K = 8.0\times 10^5\\ \ce{Cu+(aq) + 2 I- ⇌[CuI2]-} & K = 8.0\times 10^8\\ \end{array}\nonumber$

The reactions above are simplified by abbreviating the aqueous metal ion, $$\ce{[Cu(OH2)n]^+}$$ as $$\ce{Cu+(aq)}$$. This time, we are dealing with Cu(I) rather than Cu(II). Note that basicity does not seem to play a role in this series. The pKa's of the conjugate acids HCl, HBr, and HI are -6.7, -8.7, and -9.3, respectively. If basicity were important, chloride should bind most tightly, followed by bromide, and then iodide; that trend is reversed here. Instead, this case can be explained by an application of Pearson’s hard-soft acid-base theory (HSAB). HSAB theory predicts that small, charge-dense bases will bind more tightly with small, charge-dense acids. Archetypically, hard bases have N, O, or F donor atoms. Conversely, larger, more polarizable bases bind more tightly with larger, more polarizable acids. Soft bases archetypically have P or S donor atoms. In this series of copper complexes, the iodide is the softest, most polarizable of the three halide ions shown, whereas the chloride is the hardest (although not as hard as fluoride). Cu(I) has a relatively low charge density, so it can be considered a soft acid. Soft acids and soft bases form strong bonds because of the pronounced covalency between the donor and acceptor. Of the three halides, iodide is the most capable of providing that soft-soft interaction with the Cu(I) ion.

Let’s look at another major contributor to complex stability. In this case, aqueous nickel ion reacts with ammonia to produce the hexaammine complex, $$\ce{[Ni(NH3)6]^2+}$$. That’s similar to the reaction of aqueous copper ion with ammonia; ammonia is a better donor in this case than water. However, ammine is easily displaced by ethylenediamine ($$\ce{H2NCH2CH2NH2}$$, abbreviated en) to provide an ethylenediamine complex, $$\ce{[Ni(en)3]2+}$$.2,3

$\begin{array}{cl} \ce{[Ni(OH2)6]^2+ + 6 NH3 ⇌[Ni(NH3)6]^2+} & K = 2.0\times10^8 \\ \ce{[Ni(NH3)6]^2+ + 3en ⇌[Ni(en)3]^2+ + 6 NH3} & K = 4.7\times10^9\\ \end{array}\nonumber$

Once again, we see that an ammine ligand is a much more effective donor than an aquo ligand. The ethylenediamine ligand, however, binds exceptionally tightly, and can even displace ammine ligands. To gain insight into why a metal ion would have higher affinity toward $$\ce{H2NCH2CH2NH2}$$ (en) than $$\ce{NH3}$$, despite the fact that the two ligands have similar donor atoms, we can look at the thermodynamic parameters (the entropy and enthalpy) of the reaction for the formation of $$\ce{[Ni(en)3]^2+}$$ from $$\ce{[Ni(NH3)6]^2+}$$:

$\begin{array}{rcl} \Delta H^{\circ} & =& -12.1 \text{ kJ mol}^{-1}\\ \Delta S^{\circ} & = & 184.9 \text{ J mol}^{-1} K^{-1}\\ \text{and at 298 K,} -T\Delta S^{\circ} & = & -55.1 \text{ kJ mol}^{-1} \end{array}\nonumber$

That last term, $$-T\Delta S^{\circ}$$, is included for easier comparison to the enthalpy contribution, since these two factors are evaluated via the Gibbs free energy equation, $$\Delta G = \Delta H - T\Delta S$$. We see that the formation of $$\ce{[Ni(en)3]^2+}$$ from $$\ce{[Ni(NH3)6]^2+}$$ is slightly favored enthalpically (by $$\Delta H^{\circ}=-12.1 \text{kJ mol}^{-1}$$), even though there are nitrogen atom donor ligands in both reactant and product complexes. However, this small preference is not likely the most important driving force for the large formation constant. The entropy factor is much larger in comparison ($$T\Delta S^{\circ}=-55.1 \text{ kJ mol}^{-1}$$), indicating entropy is driving the huge formation constant for the ethylenediamine complex.

How do we interpret that large $$T\Delta S^{\circ}$$? The simplest approach to interpreting a change in entropy of a reaction is to evaluate the number of molecules in the reactants compared to the products. If there is an increase in the number of molecules, there is an increase in disorder (entropy).

In the replacement of aquo ligands with ammine ligands, there are seven species on either side of the equation:
$\begin{array}{rcl} \ce{[Ni(OH2)6]^2+ + 6 NH3} & \ce{⇌} & \ce{[Ni(NH3)6]^2+} \\ \text{7 molecules} & & \text{7 molecules}\\ \end{array}\nonumber$
We might not expect this reaction to have a large entropic driving force because the number of molecules is the same in the reactants and products.

On the other hand, in the replacement of ammine ligands with ethylenediamine ligands, there are four molecules on the reactant side, and seven on the products side.
$\begin{array}{rcl} \ce{[Ni(NH3)6]^2+ + 3en} & \ce{⇌} & \ce{[Ni(en)3]^2+ + 6 NH3} \\ \text{4 molecules} & & \text{7 molecules}\\ \end{array}\nonumber$

That means there is a net increase in the number of molecules in solution when the ethylenediamine replaces the ammines. An increase in the number of molecules represents an increase in the partitioning of energy, which is entropically favorable. The underlying reason for this difference is that each ethylenediamine ligand has two nitrogen donors, allowing one ethylenediamine to replace two ammines. The high formation constant for ligands that contain multiple donor atoms is called the chelate effect, from the Greek word for "crab" (which has two claws with which it can grab things). We call donors capable of binding through multiple atoms “polydentate” ligands (for “many-toothed”, indicating they can bite more strongly into the metal and hold on).

Formation constants of coordination complexes affect important processes ranging from industrial catalysis to biology. For example, most people are aware of carbon monoxide poisoning. Carbon monoxide leads to suffocation by displacing molecular oxygen from hemoglobin (Hb), the oxygen-carrying protein in blood cells. Hemoglobin has an Fe(II) ion bound in its active site which in turn coordinates molecular oxygen, increasing the oxygen-carrying capacity of a blood cell. The two relevant equilibria are:4

$\begin{array}{cl} \ce{Hb + O2 ⇌HbO2} & K = 3.2 \; \mu M^{-1} \\ \ce{Hb + CO ⇌HbCO} & K = 750 \; \mu M^{-1} \\ \end{array}\nonumber$

Here, Hb is just a shorthand for hemoglobin. The formation constants shown above indicate that carbon monoxide binds to hemoglobin hundreds of times more tightly than does oxygen. This difference can’t be caused by a chelate effect; both $$\ce{CO}$$ and $$\ce{O2}$$ are monodentate donors in this case. The origin for this difference comes from more subtle differences in metal-ligand binding that we will develop later in this chapter.

## Problems

##### Exercise $$\PageIndex{1}$$

Show how we can combine the formation constants given above for $$\ce{[Ni(NH3)6]^2+}$$ and $$\ce{[Ni(en)3]^2+}$$ from $$\ce{[Ni(HO2)6]^2+}$$ to determine the formation constant for the ethylenediamine complex from aqueous nickel ion.

$\ce{[Ni(OH2)6]^2+ + 3 en ⇄ [Ni(en)3]^2+ + 6 H2O} \qquad K = ?\nonumber$

We know the formation constants of $$\ce{[Ni(NH3)6]^2+}$$ and $$\ce{[Ni(en)3]^2+}$$. Let's call them "reactions $$i$$ and $$j$$":

$\begin{array}{lcr} \text{Reaction }i: & \ce{[Ni(OH2)6]^2+ + 6 NH3 ⇌[Ni(NH3)6]^2+} & K_i = \frac{\ce{[Ni(NH3)6]}}{\ce{[Ni(OH2)6][NH3]}} = 2.0\times10^8\\ \text{Reaction }j: & \ce{[Ni(NH3)6]^2+ + 3en ⇌[Ni(en)3]^2+ + 6 NH3} & K_j = \frac{\ce{[Ni(en)3][NH3]^6}}{\ce{[Ni(NH3)6][en]^3}}= 4.7\times10^9\\ \end{array}\nonumber$

and we want to find the equilibrium constant for the formation of $$\ce{[Ni(en)3]^2+}$$ from $$\ce{[Ni(HO2)6]^2+}$$. We'll call this "reaction $$k$$".

$\begin{array}{lcr} \text{Reaction }k: & \ce{[Ni(OH2)6]^2+ + 3 en ⇄ [Ni(en)3]^2+ + 6 H2O} & K_k = ?\\ \end{array}\nonumber$

Reaction "$$k$$" is the sum of reaction "$$i$$" followed by reaction "$$j$$".
$\begin{array}{rcl} K_k & = & K_i\times K_j \\ &=& \frac{\ce{[Ni(NH3)6]}}{\ce{[Ni(OH2)6][NH3]}} \times \frac{\ce{[Ni(en)3][NH3]^6}}{\ce{[Ni(NH3)6][en]^3}} = \frac{\ce{[Ni(en)3]}}{\ce{[Ni(OH2)6][en]^3}}\\ &=& (2.0\times 10^8 )(4.7\times 10^9 ) = 9.4\times 10^{17}\\ \end{array}\nonumber$

##### Exercise $$\PageIndex{2}$$

The formation constants shown below correspond to the substitution of one ligand on aqueous Cu(II) ion.5 Propose a trend that explains the different stabilities of the complexes.

$\begin{array}{cl} \ce{NH3} & K = 2.0 \times 10^3 \\ \ce{F-} & K = 8.0 \\ \ce{Cl-} & K = 1.2 \\ \ce{Br-} & K = 0.9 \\ \end{array}\nonumber$

This trend could be explained by basicity, since ammonium ion is a weak base and the acidity of the hydrogen halides increase in the order HF < HCl < HBr. The ammine ligand is a better donor than any of the halides to the nickel ion, but fluoride is better than chloride and chloride is better than bromide. The reason for the acidity trend varies in the midst of this series. Among the halides, fluoride is the best donor because it is the least polarizable and least stable ion. However, in the second row of the periodic table, the nitrogen in ammonia is a better donor than fluoride because nitrogen is less electronegative than fluorine. This difference is enough to compensate for the fact that ammine is a neutral ligand, presumably with less Coulombic attraction to the copper ion than fluoride.

Note that this trend is very different for Cu(II) than for a similar series of ligands with Cu(I). Cu(II), with greater charge density, is harder in character than Cu(I). Thus, it doesn’t necessarily bind more tightly to softer donors the way Cu(I) does.

##### Exercise $$\PageIndex{3}$$

Ethylenediamine can be displaced from nickel by tren, (NH2CH2CH2)3N, another polydentate ligand.3

$\ce{[Ni(en)2(OH2)2]^2+ + tren ⇄ [Ni(tren)(H2O)2]^2+ + 2 en } \qquad K = 76\nonumber$

1. What is the denticity of tren?
2. It is reported that ΔH° = +13.0 kJ mol-1 and ΔS° = 79.5 J mol-1 K-1. Propose a reason for the moderately large formation constant.

a) Tren has a denticity of 4; it is sometimes described as a tetradentate ligand.

b) The enthalpy of reaction ΔH° = +13.0 kJ mol-1 and at 298 K, -TΔS° = -23.7 kJ mol-1. That means there are no arguments based on the donor-acceptor strength, such as HSAB or basicity, because the enthalpy of reaction is positive. The tightness of binding must result from the positive entropy, which comes from the net increase in the number of molecules as the reaction proceeds (the chelate effect).

The moderately positive enthalpy change is probably related to ring strain in the multidentate complex.

## References

1. Complex-Ion Equilibria https://chem.libretexts.org/@go/page/516 (accessed Jun 25, 2021).
2. Complex Ion Formation Constants https://chem.libretexts.org/@go/page/6648 (accessed Jun 25, 2021).
3. Cotton, F.A.; Wilkinson, G. Advanced Inorganic Chemistry, 4th Ed. John Wiley & Sons: New York, 1980, pp 71-73.
4. Wilbur S, Williams M, Williams R, et al. Toxicological Profile for Carbon Monoxide. Atlanta (GA): Agency for Toxic Substances and Disease Registry (US); 2012 Jun. Table 3-12, Hemoglobin and Myoglobin Binding Kinetics and Equilibrium Constants for Oxygen and Carbon Monoxide. Available from: https://www.ncbi.nlm.nih.gov/books/NBK153687/table/T22/ (accessed Jun 26, 2021).
5. Miessler, G. L.; Fischer, P. J.; Tarr, D. A. Inorganic Chemistry, 5th Ed. Pearson: Upper Saddle River, NJ, p. 358.

This page titled 10.1.1: Thermodynamic Data is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Chris Schaller.