# Introduction to Ligand Substitution Reactions Transition metal complexes undergo various types of reactions, including redox, deprotonation, and **ligand substitution reactions**. A **ligand exchange reaction**, also known as a **ligand substitution reaction**, is a process where one ligand in a complex ion is replaced by another ligand. These reactions typically occur if the *newly formed complex is more stable* than the reacting complex. They can involve either a complete or partial exchange of ligands. This section focuses on the reactions of copper(II) and cobalt(II) complexes, involving common ligands such as water ($\text{H}_2\text{O}$), ammonia ($\text{NH}_3$), hydroxide ($\text{OH}^-$), and chloride ($\text{Cl}^-$) ions. ### i. Copper(II) Complexes Copper(II) ions in aqueous solution exist as the hexaaquacopper(II) ion, $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}(\text{aq})$, which has a characteristic blue color. #### Reaction with Hydroxide Ions ($\text{OH}^-$) When a strong base like sodium hydroxide ($\text{NaOH}$) is added to a blue solution of copper(II) sulfate ($\text{CuSO}_4$), a pale blue precipitate is observed. In this reaction, hydroxide ions effectively replace water ligands. The mechanism involves hydroxide ions acting as a base, deprotonating water molecules that are coordinated to the copper(II) ion. The resulting hydroxide ligands then substitute other water ligands. $$ \underset{\text{blue solution}}{[\text{Cu}(\text{H}_2\text{O})_6]^{2+}(\text{aq})} + 2\text{OH}^{-}(\text{aq}) \longrightarrow \underset{\text{pale blue precipitate}}{[\text{Cu}(\text{H}_2\text{O})_4(\text{OH})_2](\text{s})} + 2\text{H}_2\text{O}(\text{l}) $$ #### Reaction with Ammonia ($\text{NH}_3$) 1. **Adding aqueous ammonia directly:** If aqueous ammonia is added to the hexaaquacopper(II) ion, a pale blue precipitate is initially formed. Ammonia acts as a base, deprotonating coordinated water molecules, leading to the formation of copper(II) hydroxide. This specific reaction is often termed a **deprotonation reaction**. $$ \underset{\text{blue solution}}{[\text{Cu}(\text{H}_2\text{O})_6]^{2+}(\text{aq})} + 2\text{NH}_3(\text{aq}) \longrightarrow \underset{\text{pale blue precipitate}}{[\text{Cu}(\text{H}_2\text{O})_4(\text{OH})_2](\text{s})} + 2\text{NH}_4^+(\text{aq}) $$ 2. **Adding excess ammonia to the precipitate:** Adding *excess* ammonia solution to the pale blue precipitate of $[\text{Cu}(\text{H}_2\text{O})_4(\text{OH})_2](\text{s})$ causes it to dissolve, forming a deep blue solution of the tetraamminediaquacopper(II) ion. This is a ligand exchange reaction where ammonia ligands replace coordinated water and hydroxide ligands. $$ \underset{\text{pale blue precipitate}}{[\text{Cu}(\text{H}_2\text{O})_4(\text{OH})_2](\text{s})} + 4\text{NH}_3(\text{aq}) \longrightarrow \underset{\text{deep blue solution}}{[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}(\text{aq})} + 2\text{H}_2\text{O}(\text{l}) + 2\text{OH}^{-}(\text{aq}) $$ #### Reaction with Chloride Ions ($\text{Cl}^-$) The water ligands in the hexaaquacopper(II) ion can also be exchanged for chloride ligands by adding concentrated hydrochloric acid dropwise. This reaction results in a yellow solution of tetrachlorocuprate(II). $$ \underset{\text{blue solution}}{[\text{Cu}(\text{H}_2\text{O})_6]^{2+}(\text{aq})} + 4\text{Cl}^{-}(\text{aq}) \rightleftharpoons \underset{\text{yellow solution}}{[\text{CuCl}_4]^{2-}(\text{aq})} + 6\text{H}_2\text{O}(\text{l}) $$ This reaction is reversible. The equilibrium mixture often appears green due to the presence of both blue (reactant) and yellow (product) species. Note that the coordination number changes from 6 (octahedral) to 4 (tetrahedral) due to the larger size of chloride ligands. ### ii. Cobalt(II) Complexes Cobalt(II) ions in aqueous solution typically exist as the hexaaquacobalt(II) ion, $[\text{Co}(\text{H}_2\text{O})_6]^{2+}(\text{aq})$, which is pink. #### Reaction with Hydroxide Ions ($\text{OH}^-$) Adding a strong base like sodium hydroxide ($\text{NaOH}$) to a pink aqueous solution of cobalt(II) chloride results in a blue precipitate of cobalt(II) hydroxide. $$ \underset{\text{pink solution}}{[\text{Co}(\text{H}_2\text{O})_6]^{2+}(\text{aq})} + 2\text{OH}^{-}(\text{aq}) \longrightarrow \underset{\text{blue precipitate}}{[\text{Co}(\text{H}_2\text{O})_4(\text{OH})_2](\text{s})} + 2\text{H}_2\text{O}(\text{l}) $$ #### Reaction with Ammonia ($\text{NH}_3$) Adding concentrated aqueous ammonia slowly to the hexaaquacobalt(II) ion leads to the complete exchange of all water ligands for ammonia ligands, forming a brown solution of hexamminecobalt(II) ion. $$ \underset{\text{pink solution}}{[\text{Co}(\text{H}_2\text{O})_6]^{2+}(\text{aq})} + 6\text{NH}_3(\text{aq}) \longrightarrow \underset{\text{brown solution}}{[\text{Co}(\text{NH}_3)_6]^{2+}(\text{aq})} + 6\text{H}_2\text{O}(\text{l}) $$ #### Reaction with Chloride Ions ($\text{Cl}^-$) Addition of concentrated hydrochloric acid to an aqueous solution of cobalt(II) ions causes a ligand exchange reaction, forming a blue solution of tetrachlorocobaltate(II) complex. $$ \underset{\text{pink solution}}{[\text{Co}(\text{H}_2\text{O})_6]^{2+}(\text{aq})} + 4\text{Cl}^{-}(\text{aq}) \longrightarrow \underset{\text{blue solution}}{[\text{CoCl}_4]^{2-}(\text{aq})} + 6\text{H}_2\text{O}(\text{l}) $$ ### 6.4.1 Stability of Complexes and Stability Constant ($K_{stab}$) The **stability constant** ($K_{stab}$), also known as the formation constant, is the equilibrium constant obtained from ligand exchange reactions in a solvent. It quantifies the stability of a complex in solution. Ligand exchange reactions are reversible, allowing for the derivation of an equilibrium constant expression. The derivation is similar to that for the general equilibrium constant ($K_c$). Consider the following reversible ligand exchange reaction for cobalt(II): $$ \underset{\text{pink solution}}{[\text{Co}(\text{H}_2\text{O})_6]^{2+}(\text{aq})} + 4\text{Cl}^{-}(\text{aq}) \rightleftharpoons \underset{\text{blue solution}}{[\text{CoCl}_4]^{2-}(\text{aq})} + 6\text{H}_2\text{O}(\text{l}) $$ The position of equilibrium lies in the direction of the more stable complex. Generally, strong field ligands will substitute weak field ligands. The equilibrium constant expression ($K_c$) for this reaction would be: $$ K_c = \frac{[[\text{CoCl}_4]^{2-}][\text{H}_2\text{O}]^6}{[[\text{Co}(\text{H}_2\text{O})_6]^{2+}][\text{Cl}^-]^4} $$ Since water is the solvent and is in large excess, its concentration is considered constant and is therefore incorporated into the equilibrium constant. This modified constant is the **stability constant** ($K_{stab}$): $$ K_{stab} = \frac{[[\text{CoCl}_4]^{2-}]}{[[\text{Co}(\text{H}_2\text{O})_6]^{2+}][\text{Cl}^-]^4} $$ The units of $K_{stab}$ are calculated in the same way as those of $K_c$. However, often, the values of stability constants are very large and are reported on a $\log_{10}$ scale, in which case they are unitless. #### Factors Affecting Stability and $K_{stab}$ Values: * **Bond Strength:** Higher $K_{stab}$ values indicate that a ligand is more firmly bonded to the central metal ion. * **Electronegativity:** Ligand atoms with lower electronegativity generally form stronger bonds with the central metal ion, leading to higher stability constants. For example, ammonia ($\text{NH}_3$) forms stronger bonds than water ($\text{H}_2\text{O}$) due to nitrogen being less electronegative than oxygen when considering bonding to a metal center. * **Chelate Effect:** Complexes with polydentate ligands (ligands that can bind at multiple points) have significantly higher stability constant values than those with monodentate ligands. This phenomenon is known as the **chelate effect**. *Example:* The reaction of hexaaquacopper(II) with ethylenediamine ('en', a bidentate ligand): $$ [\text{Cu}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + 3\text{en} \longrightarrow [\text{Cu}(\text{en})_3]^{2+} + 6\text{H}_2\text{O}(\text{l}) $$ The $K_{stab}$ for the ethylenediamine complex is much higher than for monodentate ligands. #### Table of Stability Constants for Copper(II) Complexes (Log$_{10}K_{stab}$): | Complex Name | Log$_{10}K_{stab}$ | | :----------------------------------- | :----------------- | | $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$ | (Reference point, effectively low) | | $[\text{CuCl}_4]^{2-}$ | 5.6 | | $[\text{Cu}(\text{NH}_3)_6]^{2+}$ | 13.1 | | $[\text{Cu}(\text{EDTA})]^{2-}$ | 18.8 | The stability constant values are crucial for predicting which ligand substitution reactions will occur. A ligand with a higher stability constant value for a given metal ion will replace a ligand with a lower stability constant. For example, ammonia will displace chloride from tetrachlorocuprate(II) because the ammonia complex is more stable. $$ [\text{CuCl}_4]^{2-}(\text{aq}) + 6\text{NH}_3(\text{aq}) \longrightarrow [\text{Cu}(\text{NH}_3)_6]^{2+}(\text{aq}) + 4\text{Cl}^{-}(\text{aq}) $$ ## Worked Examples **Worked Example 1:** We get a mixture of aqueous solutions of tetrachlorocuprate(II) and hexaaquacopper(II) ion after adding concentrated hydrochloric acid to an aqueous solution of copper(II) sulphate. **i. Write the expression for the stability constant for the aqueous solution of tetrachlorocuprate(II).** **1. Identify the reaction:** The reaction described is: $$ [\text{Cu}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + 4\text{Cl}^{-}(\text{aq}) \rightleftharpoons [\text{CuCl}_4]^{2-}(\text{aq}) + 6\text{H}_2\text{O}(\text{l}) $$ **2. Apply the stability constant definition:** The stability constant ($K_{stab}$) includes the concentrations of product complex and reactant ligands and complex, excluding the solvent. **3. Write the expression:** $$ K_{stab} = \frac{[[\text{CuCl}_4]^{2-}]}{[[\text{Cu}(\text{H}_2\text{O})_6]^{2+}][\text{Cl}^-]^4} $$ **ii. What are the units of the stability constant?** **1. Analyze the expression for units:** The expression is $K_{stab} = \frac{[Product]}{[Reactant A][Reactant B]^4}$. The units for concentration are $\text{mol}\,dm^{-3}$. **2. Substitute units into the expression:** $$ \text{Units of } K_{stab} = \frac{(\text{mol}\,dm^{-3})}{(\text{mol}\,dm^{-3})(\text{mol}\,dm^{-3})^4} = \frac{(\text{mol}\,dm^{-3})}{(\text{mol}\,dm^{-3})^5} = (\text{mol}\,dm^{-3})^{-4} $$ **3. State the units:** The units of the stability constant ($K_{stab}$) for this reaction are $\text{mol}^{-4}\,dm^{12}$. ## Possible Questions/Answers **Concept Assessment Exercise 6.3** * **Q1:** Arrange the following ligands in increasing order of stability constants in ligand substitution reactions of iron(II) complexes: $\text{CN}^-$, $\text{OH}^-$, $\text{NH}_3$, $\text{H}_2\text{O}$, $\text{Br}^-$ **A1:** The general order of increasing ligand field strength (and often stability constant values) for common ligands is: $\text{Br}^- < \text{Cl}^- < \text{H}_2\text{O} < \text{OH}^- < \text{NH}_3 < \text{CN}^-$ Therefore, in increasing order of stability constants for iron(II) complexes: **$\text{Br}^- < \text{H}_2\text{O} < \text{OH}^- < \text{NH}_3 < \text{CN}^-$** * **Q2:** Given the stability constant, why can ammonia exchange for water in copper(II) complex? Give a reason. **A2:** Ammonia can exchange for water in copper(II) complexes because the ammonia complex, such as $[\text{Cu}(\text{NH}_3)_6]^{2+}$, has a *higher stability constant* value ($\log_{10}K_{stab} = 13.1$) compared to the aquo complex $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$. This indicates that the bond between copper(II) and ammonia is stronger than the bond between copper(II) and water, making the ammonia complex more stable. The equilibrium for the ligand exchange reaction will therefore lie towards the formation of the ammonia complex.