Which of the following divalent metal ion form the most stable complexes ?

Options

A. `Mn^(2+)`
B. `Fe^(2+)`
C. `Ni^(2+)`
D. `Cu^(2+)`

Answer

To determine which divalent metal ion forms the most stable complexes, we need to analyze the given options based on their ionic sizes and the principles of complex stability.

### Step-by-Step Solution:

1. **Understand the Concept of Complex Stability**:
- The stability of a complex ion is influenced by the charge of the metal ion and its size. Generally, a higher charge leads to greater stability. However, since we are dealing with divalent metal ions (all have a +2 charge), we need to consider their ionic sizes.

2. **Identify the Given Metal Ions**:
- The options provided are:
- Mn²⁺ (Manganese)
- Fe²⁺ (Iron)
- Ni²⁺ (Nickel)
- Cu²⁺ (Copper)

3. **Refer to the Ionic Radii**:
- The stability of the complexes is inversely related to the size of the metal ion. Smaller ions tend to form more stable complexes.
- The ionic radii (in picometers) for the divalent metal ions are approximately:
- Mn²⁺: 106 pm
- Fe²⁺: 92 pm
- Ni²⁺: 69 pm
- Cu²⁺: 73 pm

4. **Compare the Sizes**:
- From the above data, we can see that:
- Cu²⁺ has a relatively small ionic radius compared to the others.
- Ni²⁺ is also small, but larger than Cu²⁺.
- Fe²⁺ is larger than Ni²⁺.
- Mn²⁺ has the largest ionic radius among the given options.

5. **Determine the Most Stable Complex**:
- Based on the trend of ionic sizes, Cu²⁺, being the smallest, will form the most stable complex among the given divalent metal ions.

6. **Conclusion**:
- Therefore, the divalent metal ion that forms the most stable complex is **Cu²⁺**.

### Final Answer:
- The correct option is **Cu²⁺**.