Equilibrium constant for two complexes are `A: K_4[Fe(CN)_6]2.6xx10^(37)` (for dissociation) B : `K_3[Fe(CN)_6]1.9xx10^(17)` (for dissociation)

Options

A. A and B are equally stable
B. A is more stable than B
C. B is more stable than A
D. the predictable stability

Answer

To solve the problem, we need to analyze the equilibrium constants provided for the two complexes and determine which one is more stable based on the given values.

### Step-by-Step Solution:

1. **Identify the Equilibrium Constants**:
- For complex A: \( K_{A} = 2.6 \times 10^{37} \)
- For complex B: \( K_{B} = 1.9 \times 10^{17} \)

2. **Understand the Relation Between Stability and Equilibrium Constant**:
- The stability of a complex is inversely related to its equilibrium constant for dissociation. This means that a lower equilibrium constant indicates a more stable complex.
- If a complex has a high equilibrium constant, it means that it dissociates more readily, implying lower stability.

3. **Compare the Equilibrium Constants**:
- Compare \( K_{A} \) and \( K_{B} \):
- \( K_{A} = 2.6 \times 10^{37} \) (higher value)
- \( K_{B} = 1.9 \times 10^{17} \) (lower value)

4. **Determine Stability**:
- Since \( K_{B} \) is lower than \( K_{A} \), it indicates that complex B is more stable than complex A.

5. **Conclusion**:
- Therefore, the correct conclusion is that complex B is more stable than complex A.

### Final Answer:
- Complex B is more stable than complex A.