Question

Formation constants for the ammonia and ethylenediamine complexes of nickel(II) indicate that \(\mathrm{Ni}(\mathrm{en})_{3}{ }^{2+}\) is much more stable than \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) : (1) \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) \rightleftharpoons\) \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l) \quad K_{\mathrm{f}}=2.0 \times 10^{8}\) (2) \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}(a q)+3 \mathrm{en}(a q) \rightleftharpoons\) \(\mathrm{Ni}(\mathrm{en})_{3}^{2+}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l) \quad K_{\mathrm{f}}=4 \times 10^{17}\) The enthalpy changes for the two reactions, \(\Delta H_{1}^{\circ}\) and \(\Delta H_{2}^{\circ}\), should be about the same because both complexes have six Ni-N bonds. (a) Which of the two reactions should have the larger entropy change, \(\Delta S^{\circ} ?\) Explain. (b) Account for the greater stability of \(\mathrm{Ni}(\mathrm{en})_{3}^{2+}\) in terms of the relative values of \(\Delta S^{\circ}\) for the two reactions. (c) Assuming that \(\Delta H_{2}^{\circ}-\Delta H^{\circ}\) is zero, calculate the value of \(\Delta S_{2}^{\circ}-\Delta S_{1}^{\circ} .\)

Short answer

(a) \(\Delta S^{\circ}\) is larger for \(\mathrm{Ni(en)_3^{2+}}\). (b) Larger entropy change makes it more stable. (c) \(\Delta S_2^{\circ} - \Delta S_1^{\circ} > 0\).

Step-by-step solution

Analyze Reaction Stoichiometry for Entropy

For the reaction to form \(\mathrm{Ni\left(NH_3\right)_6^{2+}}\), there is a reduction in the number of moles from 7 (\(\mathrm{Ni(H_2O)_6^{2+}}\) + 6 \(\mathrm{NH_3}\)) to 7 (\(\mathrm{Ni(NH_3)_6^{2+}}\) + 6 \(\mathrm{H_2O}\)). For \(\mathrm{Ni(en)_3^{2+}}\), the change is from 4 moles (\(\mathrm{Ni(H_2O)_6^{2+}}\) + 3 en) to 7 moles (\(\mathrm{Ni(en)_3^{2+}}\) + 6 \(\mathrm{H_2O}\)). The latter results in fewer molecules from reactants to products, indicating it might have a larger positive entropy change.

Identify Reaction with Largest Entropy Change

The reaction forming \(\mathrm{Ni(en)_3^{2+}}\) leads to an increase in total moles from reactants to products comparing the release of \(\mathrm{NH_3}\). Therefore, it has a larger positive entropy change \((\Delta S_2^\circ)\).

Explain Stability through Entropy

\(\mathrm{Ni(en)_3^{2+}}\) is more stable due to a larger positive entropy change compared to \(\mathrm{Ni(NH_3)_6^{2+}}\). The large increase in disorder (\(\Delta S^{\circ}\)) makes \(K_f\) larger, leading to greater stability for \(\mathrm{Ni(en)_3^{2+}}\).

Use Given Information to Determine Entropy Difference Impact

Given \(\Delta H_2^\circ - \Delta H_1^\circ = 0\), the difference in entropy change must accommodate the difference in stability. Larger \(K_f\) in reaction 2 implies a greater \(\Delta S^{\circ}\).

Calculate \(\Delta S_2^{\circ} - \Delta S_1^{\circ}\)

Assuming \(\Delta H_2^{\circ} = \Delta H_1^{\circ}\), use \(\Delta G = \Delta H - T\Delta S\) and the relationship \(K_f = e^{-\Delta G/RT}\) to find entropy difference. Solving this gives:\[\Delta S_2^\circ - \Delta S_1^\circ = \frac{R\ln\left(\frac{K_{f2}}{K_{f1}}\right)}{T}\]With \(K_{f2} = 4 \times 10^{17}\) and \(K_{f1} = 2 \times 10^8\),\[\Delta S_2^\circ - \Delta S_1^\circ = \frac{8.314 \cdot \ln(2 \times 10^9)}{298}\].

Key concepts

Nickel(II) Complexes

Nickel(II) complexes are an intriguing class of coordination compounds where nickel is in the +2 oxidation state, coordinated to ligands. These complexes can form with various ligands, like ammonia (NH\(_3\)) or ethylenediamine (en). The stability of such complexes is often quantified by their formation constants, denoted as \(K_f\). These constants indicate the equilibrium concentration ratio of products to reactants in solution. A high \(K_f\) value implies a stable complex.
Two common complexes, \(\mathrm{Ni(NH_3)_6^{2+}}\) and \(\mathrm{Ni(en)_3^{2+}}\), exhibit different stabilities. Ethylenediamine, a bidentate ligand, forms stronger bonds with nickel due to its ability to form chelate rings, which enhances the stability of \(\mathrm{Ni(en)_3^{2+}}\) over \(\mathrm{Ni(NH_3)_6^{2+}}\). This is reflected in their respective \(K_f\) values, with \(\mathrm{Ni(en)_3^{2+}}\) having a significantly higher formation constant, indicating greater stability.
The structure and bonding in these complexes are crucial in determining their properties and potential applications in fields like catalysis and material science.

Enthalpy Changes

Enthalpy changes in a chemical reaction represent the heat absorbed or released during the reaction at constant pressure. These changes, denoted as \(\Delta H\), are a critical factor in understanding the favorability and mechanism of reactions involving nickel complexes.
When forming the nickel(II) complexes with ammonia or ethylenediamine, the enthalpy change should be similar because both complexes involve the coordination of ligands to the metal ion through Ni-N bonds. Since the number of bonds formed in both cases is the same, the energy change associated with bond formation and breaking is expected to be nearly equal, making the enthalpy changes approximately similar for these reactions.
While enthalpy is important, it's not the only factor at play. The stability difference between \(\mathrm{Ni(NH_3)_6^{2+}}\) and \(\mathrm{Ni(en)_3^{2+}}\) cannot be fully explained by \(\Delta H\), suggesting other thermodynamic aspects, such as entropy changes, also play a significant role. Understanding enthalpy provides insights into the energetic profile of the reactions but reveals only part of the broader thermodynamic picture.

Entropy Change

Entropy change, represented by \(\Delta S\), is a measure of disorder or randomness in a chemical system. In the reactions forming \(\mathrm{Ni(NH_3)_6^{2+}}\) and \(\mathrm{Ni(en)_3^{2+}}\), entropy plays a substantial role in driving the formation and stability of these complexes.
For the reaction involving ammonia, there's no net change in the number of moles, meaning the entropy change \(\Delta S_1\) is minimal. In contrast, the formation of \(\mathrm{Ni(en)_3^{2+}}\) results in a notable increase in moles from reactants to products, suggesting a larger positive entropy change \(\Delta S_2\). This increase in disorder enhances the favorability of the reaction involving ethylenediamine, contributing to the higher formation constant observed.

• The large positive \(\Delta S_2\) for \(\mathrm{Ni(en)_3^{2+}}\) amplifies the stability by boosting the attractiveness of the reaction, as seen in its high \(K_f\).
• This concept illustrates why entropy, alongside enthalpy, is crucial in comprehending the thermodynamic stability and spontaneity of reactions involving complex ion formations.

Chemical Stability

Chemical stability in coordination chemistry refers to how readily a complex remains intact without decomposing or reacting further. For nickel(II) complexes, stability is influenced by various factors, including the types of ligands, their arrangements, and thermodynamic parameters such as enthalpy and entropy.
In comparing \(\mathrm{Ni(NH_3)_6^{2+}}\) and \(\mathrm{Ni(en)_3^{2+}}\), the latter is significantly more stable. This is evident from its much larger formation constant, meaning it stays intact in solution more readily and does not dissociate easily. The stability arises from both the structure and thermodynamic changes accompanying complex formation.

• Ethylenediamine creates ring structures around the metal ion, a phenomenon known as the chelate effect, which enhances stability.
• The substantial positive entropy change further stabilizes \(\mathrm{Ni(en)_3^{2+}}\) over \(\mathrm{Ni(NH_3)_6^{2+}}\), as higher disorder drives the equilibrium towards complex formation.

The interplay of entropy and the structural advantages provided by the ligands confirms why certain complexes exhibit higher chemical stability, underpinning their practical applications in catalysis, material design, and beyond.