Iron (+II) is one of the most important oxidation states and salts are called ferrous salts. Most of the `Fe(+II)` salts are pale green and contain `[Fe(H_(2)O)_(6_]^(2+)` ion. Fe(+II) compounds are easily oxidised by air and so are difficult to obtain pure `Fe^(2+)` form many complexes like `K_(4)[Fe(CN)_(6)]. `
Q. `K_(3)[Fe(CN)_(6)]` is used in the detection of `Fe^(2+)` ion with which it gives a deep blue colour. This colour is due to the formation of :

Options

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

Answer

To solve the question regarding the detection of `Fe^(2+)` ion using `K_(3)[Fe(CN)_(6)]`, we need to understand the chemical reaction that occurs and the complex that is formed, which gives the characteristic deep blue color.

### Step-by-Step Solution:

1. **Identify the Reactants**:
We start with potassium ferricyanide, `K_(3)[Fe(CN)_(6)]`, which contains `Fe^(3+)` in its structure.

2. **Reaction with Fe(II)**:
When `K_(3)[Fe(CN)_(6)]` reacts with `Fe^(2+)`, the `Fe^(2+)` ion reduces the `Fe^(3+)` in the ferricyanide complex. The reaction can be represented as:
\[ K_3[Fe(CN)_6] + Fe^{2+} \rightarrow K_3[Fe(CN)_6] + Fe^{3+} + 2CN^- \]

3. **Formation of the Complex**:
The `Fe^(2+)` ion reacts with the `CN^-` ions to form a new complex. The product of this reaction is `Fe_3[Fe(CN)_6]_2`, which is known as Turnbull's Blue.

4. **Color Observation**:
The deep blue color observed in the solution is due to the formation of this complex, `Fe_3[Fe(CN)_6]_2`. This is a characteristic property of the complex formed when `Fe^(2+)` ions are present.

5. **Conclusion**:
Therefore, the deep blue color is due to the formation of Turnbull's Blue, which is the complex formed in the reaction between `Fe^(2+)` and `K_(3)[Fe(CN)_(6)]`.

### Final Answer:
The deep blue color observed is due to the formation of Turnbull's Blue, represented by the complex `Fe_3[Fe(CN)_6]_2`.

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