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Question

Which will form the most stable complex?

The correct answer is

C₂O₄²⁻

Understanding Complex Stability and Ligands

The stability of a complex compound is influenced by several factors, primarily related to the interaction between the central metal ion and the surrounding ligands. Ligands are ions or molecules that donate electron pairs to the metal ion to form coordinate covalent bonds. The strength of these bonds and the overall structure of the complex determine its stability.

Factors Affecting Complex Stability

Key factors that influence the stability of metal complexes include:

  • Nature of the Metal Ion: Charge density (charge/size ratio) and electron configuration play a role. Higher charge and smaller size generally lead to greater stability with anionic ligands.
  • Nature of the Ligand: This is crucial and includes several aspects:
    • Charge of the Ligand: Anionic ligands often form more stable complexes than neutral ligands with positively charged metal ions due to electrostatic attraction.
    • Basicity of the Ligand: More basic ligands (stronger electron donors) tend to form more stable complexes.
    • Denticity of the Ligand: Whether the ligand is monodentate (forms one bond), bidentate (forms two bonds), or polydentate (forms multiple bonds). Polydentate ligands that can form rings (chelates) significantly enhance stability through the chelate effect.
    • π-Bonding Ability: Some ligands can engage in π-bonding (e.g., back-bonding), which can increase complex stability (e.g., CN⁻, CO).

Analyzing the Given Ligands

Let's examine the properties of the ligands provided in the options:

Ligand Formula Charge Denticity Nature
Thiocyanate NCS<sup>⁻</sup> -1 Monodentate (can link via N or S) Anionic
Cyanide CN<sup>⁻</sup> -1 Monodentate Anionic, Strong field (π-acceptor)
Water H<sub>2</sub>O 0 Monodentate Neutral, Weak field
Oxalate C<sub>2</sub>O<sub>4</sub><sup>2-</sup> -2 Bidentate Anionic, Chelating

The Chelate Effect and Stability

The chelate effect is a phenomenon where complexes formed by chelating ligands (polydentate ligands) are significantly more stable than complexes formed by comparable monodentate ligands. This increased stability is primarily due to a favorable entropy change during complex formation. When a chelating ligand binds to a metal ion, it replaces several solvent molecules or monodentate ligands. For example, a bidentate ligand replaces two monodentate ligands. The formation of the complex involves one molecule of the bidentate ligand replacing two separate molecules of the monodentate ligand. This typically leads to an increase in the number of species in solution (metal complex + multiple released solvent/monodentate molecules vs. metal ion + multiple monodentate ligands), which increases entropy and drives the reaction towards complex formation, resulting in higher stability.

Let's consider a metal ion $\text{M}^{n+}$ forming a complex with monodentate ligands $\text{L}$ and a bidentate ligand $\text{B}$ (which occupies two coordination sites).

Reaction with monodentate ligand (e.g., ethylamine, $\text{EtNH}_2$, comparable basicity to ethylenediamine):

$\text{[M(H}_2\text{O)}_6\text{]}^{n+} + 2\text{EtNH}_2 \rightleftharpoons \text{[M(H}_2\text{O)}_4\text{(EtNH}_2\text{)}_2\text{]}^{n+} + 2\text{H}_2\text{O}$

Reaction with bidentate ligand (e.g., ethylenediamine, en):

$\text{[M(H}_2\text{O)}_6\text{]}^{n+} + \text{en} \rightleftharpoons \text{[M(H}_2\text{O)}_4\text{(en)]}^{n+} + 2\text{H}_2\text{O}$

Even if the enthalpy change ($\Delta H$) for these reactions might be similar (considering the strength of the bonds formed), the entropy change ($\Delta S$) is significantly more positive for the reaction with the bidentate ligand because one molecule (en) replaces two molecules (2 H$_2$O or 2 EtNH$_2$). Since $\Delta G = \Delta H - T\Delta S$, a more positive $\Delta S$ leads to a more negative $\Delta G$, indicating a more stable complex (larger formation constant $K$).

Comparing Ligands and Determining Stability

Among the given options:

  • H₂O is a neutral, monodentate ligand and generally forms less stable complexes compared to anionic or chelating ligands.
  • NCS⁻ and CN⁻ are anionic, monodentate ligands. CN⁻ is known to be a very strong field ligand, often forming stable complexes, partly due to π-bonding effects. NCS⁻ is also anionic.
  • C₂O₄²⁻ (oxalate) is an anionic, bidentate ligand. It forms a five-membered ring with the metal ion, leading to the chelate effect.

While CN⁻ is a strong ligand in terms of crystal field splitting, the bidentate nature of oxalate and the resulting chelate effect typically confer greater overall thermodynamic stability (higher formation constant) compared to monodentate ligands, even strong ones like CN⁻, when comparing complexes of similar coordination numbers and charges.

Therefore, the chelation effect provided by the bidentate oxalate ligand (C₂O₄²⁻) makes the complex formed with it significantly more stable than those formed by the monodentate ligands (NCS⁻, CN⁻, H₂O).

Conclusion on Most Stable Complex

Considering the factors, especially the dominant chelate effect, the ligand that will form the most stable complex among the given options is C₂O₄²⁻.

Revision Table: Complex Stability Factors

Factor Influence on Stability Example (among options)
Ligand Charge Anionic > Neutral (with positive metal ions) NCS⁻, CN⁻, C₂O₄²⁻ are anionic (favored over H₂O)
Ligand Denticity Bidentate > Monodentate (Chelate Effect) C₂O₄²⁻ is bidentate (favored over NCS⁻, CN⁻, H₂O)
Ligand Field Strength Strong field often > Weak field CN⁻ is very strong field; C₂O₄²⁻ is strong field; H₂O is weak field

Additional Information: Formation Constants

Complex stability is quantitatively measured by its formation constant (or stability constant), denoted by K. This constant is the equilibrium constant for the formation of the complex from the metal ion and ligands in solution. A higher formation constant indicates a more stable complex.

For a stepwise formation:

$\text{M} + \text{L} \rightleftharpoons \text{ML} \quad K_1 = \frac{\text{[ML]}}{\text{[M][L]}}$

$\text{ML} + \text{L} \rightleftharpoons \text{ML}_2 \quad K_2 = \frac{\text{[ML}_2\text{]}}{\text{[ML][L]}}$

and so on.

The overall formation constant ($\beta_n$) for a complex $\text{ML}_n$ is the product of the stepwise constants:

$\beta_n = K_1 \times K_2 \times \dots \times K_n = \frac{\text{[ML}_n\text{]}}{\text{[M][L]}^n}$

Comparing the β values for complexes of a specific metal ion with different ligands provides a direct measure of their relative stability. Studies consistently show that complexes with chelating ligands have significantly higher formation constants compared to those with monodentate ligands, even if the monodentate ligands are strong field or highly basic, demonstrating the power of the chelate effect in enhancing complex stability.

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Important Questions from Coordination Compounds

  1. Match List-I with List-II:

    List-IList-II
    (A) Diamagnetic solid(I) CrO₂
    (B) Ferromagnetic solid(II) Fe₃O₄
    (C) Antiferromagnetic solid(III) NaCl
    (D) Ferrimagnetic solid(IV) MnO

    Choose the correct answer from the options given below:

  2. [NiCl₂(PPh₃)₂] is named as:

  3. Inner orbital complex among the following is:

    (A) [Co(NH₃)₆]³⁺

    (B) [CoF₆]³⁻

    (C) [Ni(CN)4]²⁻

    (D) [MnCl₆]³⁻

    (E) [FeF₆]³⁻

    Choose the correct answer from the options given below:

  4. How many Cr-O bonds in dichromate ions are of the same bond length and are in resonance?

  5. Which of the following statement is/are correct for complex [NiCl4]2-?

    (A) Ni has oxidation state +2

    (B) Cl is a weak field ligand

    (C) Compound is paramagnetic

    (D) dsp2 hybridisation

    (E) Low spin complex

    Choose the correct answer from the options given below:

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