A metal ion give chocolate coloured ppt with `K_(4)[Fe(CN)_(6)]`
What is the oxidation state of that metal in its ion ?

Answer

To determine the oxidation state of the metal ion that gives a chocolate-colored precipitate with potassium ferrocyanide (K₄[Fe(CN)₆]), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Metal Ion**: The question states that the metal ion gives a chocolate-colored precipitate with K₄[Fe(CN)₆]. Common metal ions that form colored precipitates with potassium ferrocyanide include copper (Cu²⁺), iron (Fe²⁺, Fe³⁺), and zinc (Zn²⁺).

2. **Analyze the Precipitate Color**: The color of the precipitate is a crucial clue. The chocolate brown precipitate is characteristic of the reaction between potassium ferrocyanide and copper(II) ions (Cu²⁺).

3. **Write the Reaction**: When copper(II) sulfate (CuSO₄) reacts with potassium ferrocyanide, the reaction can be represented as:
\[ \text{CuSO}_4 + \text{K}_4[\text{Fe(CN)}_6] \rightarrow \text{Cu}_2[\text{Fe(CN)}_6] + \text{K}_2\text{SO}_4 \]
This reaction indicates the formation of a precipitate of copper ferrocyanide.

4. **Determine the Oxidation State of Copper**: In the compound CuSO₄, the sulfate ion (SO₄²⁻) has a charge of -2. To balance this charge, the oxidation state of copper (Cu) must be +2. Therefore, we can set up the equation:
\[ X + (-2) = 0 \implies X = +2 \]
where X is the oxidation state of copper.

5. **Conclusion**: Since the metal ion that gives the chocolate-colored precipitate with K₄[Fe(CN)₆] is copper, and we have determined its oxidation state to be +2, we conclude that the oxidation state of the metal ion is +2.

### Final Answer:
The oxidation state of the metal ion is +2.

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