What product is formed by mixing the solution of `K_4 [Fe (CN)_6]` with the solution of `FeCl_2`?

Options

A. Ferro ferricyanide
B. Ferric ferrocyanide
C. Ferric ferricyanide
D. None

Answer

To determine the product formed by mixing the solution of \( K_4[Fe(CN)_6] \) (potassium ferrocyanide) with the solution of \( FeCl_2 \) (iron(II) chloride), we can follow these steps:

### Step 1: Identify the Reactants
We have two reactants:
1. \( K_4[Fe(CN)_6] \) - This is known as potassium ferrocyanide.
2. \( FeCl_2 \) - This is iron(II) chloride.

### Step 2: Understand the Reaction
When potassium ferrocyanide reacts with iron(II) chloride, it leads to the formation of a complex compound. Ferrocyanide ions (\( [Fe(CN)_6]^{4-} \)) can react with iron(II) ions (\( Fe^{2+} \)) to form a precipitate.

### Step 3: Write the Reaction
The reaction can be represented as:
\[ K_4[Fe(CN)_6] + 2FeCl_2 \rightarrow 2KCl + [Fe_2[Fe(CN)_6]] \]
Here, \( [Fe_2[Fe(CN)_6]] \) is the complex formed, which is known as Turnbull's blue.

### Step 4: Identify the Product
The product formed is \( [Fe_2[Fe(CN)_6]] \), which is a deep blue precipitate known as Turnbull's blue.

### Final Answer
The product formed by mixing the solution of \( K_4[Fe(CN)_6] \) with the solution of \( FeCl_2 \) is Turnbull's blue, represented as \( [Fe_2[Fe(CN)_6]] \).

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