Question

Identify the correct decreasing order of stability of complexes formed by divalent metal ions with same ligand.

• A. $\text{Cu}^{2+} > \text{Mn}^{2+} > \text{Cd}^{2+}$
• B. $\text{Cd}^{2+} > \text{Mn}^{2+} > \text{Cu}^{2+}$
• C. $\text{Mn}^{2+} > \text{Cd}^{2+} > \text{Cu}^{2+}$
• D. $\text{Cu}^{2+} > \text{Cd}^{2+} > \text{Mn}^{2+}$

Hint:

Remember: Copper ($Cu^{2+}$) is almost always at the top of the stability chain for divalent transition metals.

Answer 1

The Correct Option is D

Step 1: Concept
The stability of complexes for divalent metal ions of the first transition series follows the Irving-Williams order.

Step 2: Meaning
Stability increases as the ionic radius decreases and effective nuclear charge increases across the period.

Step 3: Analysis
The Irving-Williams order is: $\text{Ba}^{2+} < \text{Sr}^{2+} < \text{Ca}^{2+} < \text{Mg}^{2+} < \text{Mn}^{2+} < \text{Fe}^{2+} < \text{Co}^{2+} < \text{Ni}^{2+} < \text{Cu}^{2+} > \text{Zn}^{2+}$. $\text{Cu}^{2+}$ typically forms the most stable complexes due to Jahn-Teller distortion and smaller size.

Step 4: Conclusion
Comparing the given ions, $\text{Cu}^{2+}$ is the most stable and $\text{Mn}^{2+}$ is the least stable among them. Final Answer: (D)