Which will form the most stable complex?
C₂O₄²⁻
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.
Key factors that influence the stability of metal complexes include:
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 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 $K$).
Among the given options:
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).
Considering the factors, especially the dominant chelate effect, the ligand that will form the most stable complex among the given options is C₂O₄²⁻.
| 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 |
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.
Match List-I with List-II:
| List-I | List-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:
[NiCl₂(PPh₃)₂] is named as:
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:
How many Cr-O bonds in dichromate ions are of the same bond length and are in resonance?
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: