Question

Which from the following cations forms the least stable complex with the same ligand?

• A. \( \text{Co}^{2+} \)
• B. \( \text{Fe}^{2+} \)
• C. \( \text{Cd}^{2+} \)
• D. \( \text{Cu}^{2+} \)

Hint:

In transition metal complexes, the size and charge of the metal ion significantly influence the stability of the complex. A higher charge-to-radius ratio typically leads to greater stability.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The stability of coordination complexes depends on factors like the charge and radius of the metal cation. For divalent 3d transition metals, the stability follows the Irving-Williams series.
Step 2: Detailed Explanation:
The Irving-Williams series for divalent metal complex stability is:
\( \text{Mn}^{2+}<\text{Fe}^{2+}<\text{Co}^{2+}<\text{Ni}^{2+}<\text{Cu}^{2+}>\text{Zn}^{2+} \).
- \( \text{Cu}^{2+} \) forms the most stable complexes among these 3d metals.
- \( \text{Fe}^{2+} \) and \( \text{Co}^{2+} \) are in the middle of the series.
- \( \text{Cd}^{2+} \) is a 4d metal ion. It is significantly larger than the 3d transition metal ions listed. Due to its larger ionic radius and lower charge density, its complexes with typical N/O donor ligands are generally less stable than those of the 3d transition series of similar charge.
Step 3: Final Answer:
\( \text{Cd}^{2+} \) forms the least stable complex among the given options.