The oxidation state of Fe in `K_(4)[Fe(CN)_(6)]` is

Options

A. `+2`
B. `+6`
C. `+3`
D. `+4`

Explanation

To determine the oxidation state of iron (Fe) in the compound \( K_4[Fe(CN)_6] \), we can follow these steps:

### Step 1: Identify the components of the compound
The compound consists of potassium ions (K\(^+\)) and the complex ion \([Fe(CN)_6]^{4-}\).

### Step 2: Assign known oxidation states
- Potassium (K) is an alkali metal and has an oxidation state of +1.
- The cyanide ion (CN\(^-\)) is a ligand and has an oxidation state of -1.

### Step 3: Set up the equation
Let the oxidation state of iron (Fe) be \( x \). The complex ion \([Fe(CN)_6]\) has a total charge of -4. We can express this relationship as follows:

\[ \text{Total charge} = \text{oxidation state of K} + \text{oxidation state of Fe} + \text{oxidation state of CN} \]

The equation can be set up as:

\[ 4(+1) + x + 6(-1) = 0 \]

### Step 4: Substitute known values
Substituting the known values into the equation gives us:

\[ 4(1) + x + 6(-1) = 0 \]

This simplifies to:

\[ 4 + x - 6 = 0 \]

### Step 5: Solve for \( x \)
Now, we can solve for \( x \):

\[ x - 2 = 0 \]

\[ x = +2 \]

### Conclusion
The oxidation state of iron (Fe) in \( K_4[Fe(CN)_6] \) is +2.

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