Oxidation state of Fe in `K_(4)[Fe(CN)_(6)]`

Answer

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

### Step 1: Assign variables to known oxidation states
Let the oxidation state of iron (Fe) be \( x \). We know the oxidation states of the other components in the complex:
- Potassium (K) has an oxidation state of \( +1 \).
- The cyanide ion (CN) has an oxidation state of \( -1 \).

### Step 2: Write the equation based on the overall charge
The overall charge of the complex ion \( [Fe(CN)_6]^{4-} \) is \( 0 \) because it is neutral when combined with the 4 potassium ions. Therefore, we can set up the equation:

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

### Step 3: Substitute known values into the equation
Substituting the known values into the equation gives us:

\[ 4 + x - 6 = 0 \]

### Step 4: Simplify the equation
Now, simplify the equation:

\[ x - 2 = 0 \]

### Step 5: Solve for \( x \)
Solving for \( x \) gives:

\[ x = +2 \]

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

---