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Consider the reactions $$\mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) \longrightarrow \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q)$$ $$\mathrm{Ni}^{2+}(a q)+3 \mathrm{en}(a q) \longrightarrow \mathrm{Ni}(\mathrm{en})_{3}^{2+}(a q)$$ where $$\mathrm{en}=\mathrm{H}_{2} \mathrm{N}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2}$$ The \(\Delta H\) values for the two reactions are quite similar, yet \(\mathrm{K}_{\text { reaction } 2}>K_{\text { reaction }}\) . Explain.

Short Answer

Expert verified
In summary, the difference in stability between the two coordination complexes is due to the stronger interaction between the bidentate ethylenediamine ligand and the nickel ion compared to the monodentate ammonia ligand. This results in a higher equilibrium constant (\(K\)) for reaction 2, as the more stable complex \(Ni(en)_3^{2+}\) is favored compared to the less stable complex \(Ni(NH_3)_6^{2+}\).

Step by step solution

01

Identify the differences between the ligands in the two reactions

The first reaction involves forming a coordination complex with ammonia \(NH_3\) ligands whereas the second reaction involves forming a coordination complex with ethylenediamine \(en\) ligands. The main difference between the two ligands is their structure, with ammonia being a monodentate ligand and ethylenediamine being a bidentate ligand.
02

Explain the significance of bidentate ligands

Bidentate ligands like ethylenediamine can form two coordinate bonds with the central metal ion, in this case nickel. This leads to a more stable complex, as the bidentate ligand can effectively hold the metal ion in place through two bonds, resulting in a stronger interaction between the metal ion and the ligand.
03

Discuss the difference in stability between the coordination complexes

Since the complex formed with ethylenediamine is more stable due to the stronger interaction between the metal ion and the ligand, the equilibrium for reaction 2 will favor the formation of the \(Ni(en)_3^{2+}\) complex. In contrast, the \(Ni(NH_3)_6^{2+}\) complex formed with ammonia ligands is less stable, due to the weaker interaction between the metal ion and the ligand.
04

Relate the difference in stability to equilibrium constants

The greater the stability of the coordination complex, the more the equilibrium will favor its formation. In other words, the equilibrium constant for the formation of the complex will be higher when the complex is more stable. In this case, since the \(Ni(en)_3^{2+}\) complex is more stable than the \(Ni(NH_3)_6^{2+}\) complex, the equilibrium constant (\(K\)) for reaction 2 will be greater than the equilibrium constant for reaction 1.
05

Conclude

In summary, the equilibrium constant \(K\) for reaction 2 is greater than the equilibrium constant for reaction 1 because the ethylenediamine ligand forms a more stable coordination complex with the nickel ion compared to the ammonia ligand. This leads to a stronger interaction between the metal ion and the ligand, resulting in a higher equilibrium constant for the reaction involving ethylenediamine.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bidentate Ligands
Bidentate ligands play a crucial role in coordination chemistry as they can form two bonds with a central metal ion. A classic example is ethylenediamine (en), which has two nitrogen atoms, each with a lone pair of electrons ready to bond with a metal ion.
This dual bonding capability allows bidentate ligands to create more stable complexes by effectively "holding onto" the metal ion at two points. Imagine a situation where holding an object with both hands provides more security and stability. Similarly, bidentate ligands achieve this with metal ions.
For nickel in this case, ethylenediamine's two binding sites wrap around the metal ion tightly, ensuring a more stable association.
This increased stability is why bidentate ligands like ethylenediamine often form more stable complexes as compared to monodentate ligands like ammonia.
Complex Stability
Complex stability refers to the persistence of a coordination complex in a solution. In the context of nickel complexes, the stability can be compared between the complexes of nickel with ammonia and ethylenediamine. Complexes with bidentate (or even multidentate) ligands, such as ethylenediamine, tend to be more stable. Here are a few reasons why:
  • **Chelate Effect**: When a bidentate ligand forms a complex, it's known as a chelate complex, where the ligand wraps around the metal ion almost like a claw.
  • **Entropy Factor**: Forming fewer but more tightly bonded chelate rings often increases the disorder (entropy) of the system, which generally favors more stable complexes.
The enhanced stability of these chelate complexes results in a greater persistence in solution, thereby favoring their formation when compared to complexes with monodentate ligands, like ammonia.
Equilibrium Constants
In coordination chemistry, equilibrium constants (\(K\)) express the degree to which a reaction favors the formation of products. Specifically, they quantify the "forward" progress of forming complexes from their simpler precursors. Let's examine the nickel reactions mentioned earlier:
- Reaction 1 with ammonia results in \(Ni(NH_3)_6^{2+}\).- Reaction 2 with ethylenediamine leads to \(Ni(en)_3^{2+}\).Since ethylenediamine forms a stronger interaction, it naturally drives the equilibrium towards the formation of \(Ni(en)_3^{2+}\), which implies a higher equilibrium constant compared to the ammonia reaction.
Therefore, a greater equilibrium constant in the second reaction indicates a preference for forming the more stable ethylenediamine complex.
This concept shows how coordination chemistry often revolves around equilibriums and the preferences dictated by both ligands and their interactions.
Ligand Interactions
Ligand interactions are foundational to understanding coordination complexes. These interactions involve the bonding between metal ions and the molecules or ions surrounding them, known as ligands. Different ligands can bond to metal ions with varying strengths and structures. Here's a closer look:
  • **Monodentate Ligands**: Can only form one bond with a metal ion.
    Ammonia is an example, and its weaker bonding results in less stable complexes.
  • **Bidentate Ligands**: Form two bonds with a metal ion, providing considerable stability.
    Ethylenediamine is a prime example.
  • **Interaction Strength**: The number of bonds and the nature of the ligand strongly influence the strength of the interaction.
    Strong interactions often lead to more stable complexes and, consequently, larger equilibrium constants.
Understanding these interactions helps not only in predicting the stability of complexes but also in manipulating chemical reactions to favor certain products. It's a key principle in rationalizing why some complexes form more readily than others in coordination chemistry.

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Most popular questions from this chapter

The following reaction occurs in pure water: $$\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)$$ which is often abbreviated as $$\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)$$ For this reaction, \(\Delta G^{\circ}=79.9 \mathrm{kJ} / \mathrm{mol}\) at \(25^{\circ} \mathrm{C}\) . Calculate the value of \(\Delta G\) for this reaction at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{OH}^{-}\right]=0.15 M\) and \(\left[\mathrm{H}^{+}\right]=0.71 M .\)

Consider the reaction: $$\mathrm{H}_{2} \mathrm{S}(g)+\mathrm{SO}_{2}(g) \longrightarrow 3 \mathrm{S}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ for which \(\Delta H\) is \(-233 \mathrm{kJ}\) and \(\Delta S\) is \(-424 \mathrm{J} / \mathrm{K}\) . a. Calculate the free energy change for the reaction \((\Delta G)\) at 393 \(\mathrm{K} .\) b. Assuming \(\Delta H\) and \(\Delta S\) do not depend on temperature, at what temperatures is this reaction spontaneous?

Carbon tetrachloride \(\left(\mathrm{CCl}_{4}\right)\) and benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) form ideal solutions. Consider an equimolar solution of \(\mathrm{CC}_{4}\) and \(\mathrm{C}_{6} \mathrm{H}_{6}\) at \(25^{\circ} \mathrm{C}\) . The vapor above the solution is collected and condensed. Using the following data, determine the composition in mole fraction of the condensed vapor.

For each of the following pairs of substances, which substance has the greater value of \(S^{\circ} ?\) a. \(\mathrm{C}_{\text { graphite }}(s)\) or \(\mathrm{C}_{\text { diamond }}(s)\) b. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) or \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)\) c. \(\mathrm{CO}_{2}(s)\) or \(\mathrm{CO}_{2}(g)\)

Calculate the entropy change for the vaporization of liquid methane and liquid hexane using the following data. $$\begin{array}{lll} {\text { }} & {\text { Boiling Point (1 atm)}} & { \Delta H_{\mathrm{vap}} } \\ \hline {\text { Methane }} & \quad\quad\quad {112 \mathrm{K}} & {8.20 \mathrm{kJ} / \mathrm{mol}} \\ {\text { Hexane }} & \quad\quad\quad {342 \mathrm{K}} & {28.9 \mathrm{kJ} / \mathrm{mol}}\end{array}$$ Compare the molar volume of gaseous methane at 112 \(\mathrm{K}\) with that of gaseous hexane at 342 \(\mathrm{K}\) . How do the differences in molar volume affect the values of \(\Delta S_{\mathrm{vap}}\) for these liquids?

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