Turnbull's blue is formed when `Fe^(+2)` ions are added to `K_(3)[Fe(CN)_(6)]` Turnbull's blue is

Options

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

Answer

To determine what Turnbull's blue is formed from the reaction of `Fe^(+2)` ions with `K_(3)[Fe(CN)_(6)]`, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants in this reaction are `Fe^(+2)` ions and potassium ferricyanide, which is represented as `K3[Fe(CN)6]`.

2. **Understand the Reaction**:
- When `Fe^(+2)` ions are added to potassium ferricyanide, a chemical reaction occurs. The `Fe^(+2)` ions react with the ferricyanide ions present in the potassium ferricyanide.

3. **Write the Reaction**:
- The reaction can be represented as:
\[ 3Fe^{2+} + 2K_3[Fe(CN)_6] \rightarrow Fe_3[Fe(CN)_6]_2 + 6K^+ \]
- Here, `Fe_3[Fe(CN)_6]_2` is the product formed, which is known as Turnbull's blue.

4. **Identify the Product**:
- The product formed, `Fe_3[Fe(CN)_6]_2`, is the compound known as Turnbull's blue. It is a blue precipitate that forms during this reaction.

5. **Conclusion**:
- Therefore, Turnbull's blue is identified as `Fe_3[Fe(CN)_6]_2`, which is also referred to as ferrous ferricyanide.

### Final Answer:
Turnbull's blue is `Fe_3[Fe(CN)_6]_2`.

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