The oxidation number of iron in `K_4[Fe(CN)_6]` is

Options

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

Explanation

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

### Step 1: Identify the oxidation states of known elements
In the compound \( K_4[Fe(CN)_6] \):
- Potassium (K) has an oxidation state of +1. Since there are 4 potassium atoms, their total contribution to the oxidation state is:
\[ 4 \times (+1) = +4 \]

### Step 2: Determine the oxidation state of the cyanide ion (CN)
The cyanide ion (CN) has a known oxidation state of -1. Since there are 6 cyanide ions in the compound, their total contribution to the oxidation state is:
\[ 6 \times (-1) = -6 \]

### Step 3: Set up the equation for the oxidation state of iron (Fe)
Let the oxidation state of iron be \( x \). The overall charge of the complex ion \( [Fe(CN)_6] \) is neutral (0). Therefore, we can set up the following equation:
\[ x + (+4) + (-6) = 0 \]

### Step 4: Solve for the oxidation state of iron (Fe)
Now, simplify the equation:
\[ x + 4 - 6 = 0 \]
\[ x - 2 = 0 \]
\[ x = +2 \]

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

### Final Answer
The oxidation number of iron in \( K_4[Fe(CN)_6] \) is +2.