Problem 90 - Chemistry and Chemical Reactivity

Question

In this question, we explore the differences between metal coordination by monodentate and bidentate ligands. Formation constants, \(K_{t}\), for \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}(\mathrm{aq})\) and \(\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}(\mathrm{aq})\) are as follows: \(\mathrm{Ni}^{2+}(\mathrm{aq})+6 \mathrm{NH}_{3}(\mathrm{aq}) \longrightarrow\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}(\mathrm{aq}) \quad K_{\mathrm{f}}=10^{8}\) \(\mathrm{Ni}^{2+}(\mathrm{aq})+3 \mathrm{en}(\mathrm{aq}) \longrightarrow\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}(\mathrm{aq})\) \(K_{f}=10^{18}\) The difference in \(K_{f}\) between these complexes indicates a higher thermodynamic stability for the chelated complex, caused by the chelate effect. Recall that \(K\) is related to the standard free energy of the reaction by \(\Delta_{r} G^{\circ}=-R T \ln K\) and \(\Delta_{r} G^{\circ}=\) \(\Delta_{r} H^{\circ}-T \Delta_{r} S^{\circ} .\) We know from experiment that \(\Delta_{t} H^{\circ}\) for the \(\mathrm{NH}_{3}\) reaction is \(-109 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn}\) and \(\Delta_{i} H^{\circ}\) for the ethylenediamine reaction is \(-117 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn} .\) Is the difference in \(\Delta_{r} H^{\circ}\) suffi- cient to account for the \(10^{10}\) difference in \(K_{f} ?\) Comment on the role of entropy in the second reaction.

Step-by-Step Solution

Answer

The small enthalpy difference cannot account for the large \( K_f \) difference. Entropy plays a key role in chelate stability.

Step 1: Understand the Problem

We need to determine whether the difference in enthalpy change \( \Delta_r H^\circ \) is sufficient to account for the significant difference in formation constants \( K_f \) between two nickel complexes, one with monodentate ligands \( \text{NH}_3 \) and one with bidentate ligands (ethylenediamine, \( \text{en} \)). Additionally, we should consider the role of entropy in the higher stability of the chelated complex.

Step 2: Enthalpy Considerations

The difference in enthalpy \( \Delta_r H^\circ \) for the two reactions is calculated as follows: \( \Delta_r H^\circ (\mathrm{en}) = -117 \mathrm{kJ/mol} \) and \( \Delta_r H^\circ (\text{NH}_3) = -109 \mathrm{kJ/mol} \). The difference is \(-117 - (-109) = -8 \mathrm{kJ/mol}\). This is a small difference, suggesting that enthalpy alone is unlikely to account for a \( 10^{10} \) difference in \( K_f \).

Step 3: The Role of Entropy

The large difference in \( K_f \) suggests that entropy \( \Delta_r S^\circ \) plays a major role in stabilizing the ethylenediamine complex. Using the equation \( \Delta_r G^\circ = \Delta_r H^\circ - T \Delta_r S^\circ \) and given \( \Delta_r G^\circ = -RT \ln K \), we can deduce that for the bidentate ligand, the chelate effect significantly increases entropy, leading to a more negative \( \Delta_r G^\circ \). This increase in entropy happens because fewer molecules are present in the system after complexation, minimizing disorder.

Key Concepts

Monodentate Ligands

Monodentate ligands are molecules or ions that form a single bond with a central metal ion. Imagine them as having one claw that holds onto the metal. This single point of attachment is like a tether that binds them to the metal center.

For instance, here we have ammonia (\(\text{NH}_3\)) acting as a monodentate ligand. It coordinates through its lone pair of electrons on the nitrogen atom to bond with a nickel ion (\(\text{Ni}^{2+}\)). Monodentate ligands tend to make the complex more straightforward in structure, but sometimes less stable thermodynamically compared to their multidentate counterparts, due to the weaker interaction. This can result in flexibility and possibly even dissociation from the metal center.

• This type of complex can easily exchange ligands.
• Typically results in lower formation constants.
• Is often seen in simpler molecular geometries.

Bidentate Ligands

Bidentate ligands play an essential role by offering two points of attachment to a metal center, almost like two arms wrapping around the metal. This dual-attachment mechanism offers increased stability due to the chelate effect.

Ethylenediamine (\(\text{en}\)) is a prime example of a bidentate ligand. It has two nitrogen atoms, each can donate a pair of electrons, forming a stronger and more stable complex with the nickel ion (\(\text{Ni}^{2+}\)). This double-arm embrace creates five or six-membered rings with the metal, which are energetically favored.

• Enhances the overall stability of the complex.
• Increases the entropy upon complex formation, often driving the reaction to completion
• Can significantly influence the geometry and properties of the complexed metal ion.

Formation Constants

Formation constants (\(K_f\)) are crucial for understanding the stability of metal complexes. They represent the equilibrium constant for the formation of a complex ion from the metal ion and its ligands in a solution. A higher \(K_f\) value means a more stable complex.

In the given example, the formation constant for the ammonia complex is \(10^8\), while for ethylenediamine, it is \(10^{18}\). This vast difference indicates that the bidentate ligand (\(\text{en}\)) forms a much more stable complex than the monodentate ligand (\(\text{NH}_3\)). The chelate effect plays a significant role here, where multiple bonds formed by bidentate ligands lead to more thermodynamically stable complexes.

• \(K_f\) is directly related to the free energy change in the reaction.
• The higher the \(K_f\), the more negative the standard Gibbs free energy change.
• This stability is primarily due to the increased entropy and enthalpy considerations.