Consider the following experiments and answer the questions at the end of it
(A) When `Fe(CN)_(2)` solution is treated with `KCN` solution species formed no longer gives tests of `Fe^(2+)` and `CN^(Θ)`
(B) When `K_(2)SO_(4)` solution is treated with `A1_(2)(SO_(4))_(3)` solution species formed gives tests of `K^(o+),A1^(3+)` and `SO_(4)^(2-)`
Species formed in experiment `A` does not give test of `Fe^(2+)` and `CN^(Θ)` it is due to formation of .

Options

A. `K_(2)[Fe(CN)_(4)]`
B. `K_(3)[Fe(CN)_(5)]`
C. `K_(3)[Fe(CN)_(6)]`
D. `K_(4)[Fe(CN)_(6)]`

Explanation

To solve the problem, we need to analyze the two experiments provided and understand the chemical reactions taking place.

### Step-by-Step Solution:

1. **Understanding Experiment A**:
- We start with a solution of `Fe(CN)2` and treat it with `KCN`.
- The problem states that the species formed no longer gives tests for `Fe^(2+)` and `CN^(-)`.
- This indicates that both the iron and cyanide ions are no longer free in the solution.

2. **Formation of a Complex**:
- The reason for the absence of tests for `Fe^(2+)` and `CN^(-)` is that they have likely formed a complex.
- When `Fe(CN)2` reacts with `KCN`, a complex compound is formed, specifically `K4[Fe(CN)6]`, which is known as potassium ferrocyanide.

3. **Coordination Sphere**:
- In the complex `K4[Fe(CN)6]`, the `Fe^(2+)` ion is coordinated to six cyanide ions (`CN^(-)`), forming a stable coordination complex.
- The coordination sphere consists of the central metal ion (Fe) and the ligands (CN).

4. **Testing for Ions**:
- Since `Fe^(2+)` and `CN^(-)` are now part of the coordination complex, they cannot be detected individually in the solution.
- However, potassium ions (`K^+`) can still be detected since they are outside the coordination sphere.

5. **Conclusion**:
- The species formed in Experiment A does not give tests for `Fe^(2+)` and `CN^(-)` due to the formation of the coordination complex `K4[Fe(CN)6]`.

### Final Answer:
The species formed in Experiment A is due to the formation of the complex `K4[Fe(CN)6]`.

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