O.N. (Oxidation Number) of Fe in `K_(4)[Fe(CN)_(6)]` is

Options

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

Explanation

To find the oxidation number (O.N.) of 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\(^+\)), iron (Fe), and cyanide ions (CN\(^-\)).

### Step 2: Determine the oxidation state of known ions
- Potassium (K) is an alkali metal and has a fixed oxidation state of +1.
- The cyanide ion (CN) has a known oxidation state of -1.

### Step 3: Set up the equation
Let the oxidation number of Fe be \( x \). The overall charge of the complex ion \( [Fe(CN)_6]^{4-} \) must equal the total charge contributed by the potassium ions and the cyanide ions.

The equation can be set up as follows:
\[ 4(+1) + x + 6(-1) = 0 \]

### Step 4: Simplify the equation
Calculating the contributions:
- The total contribution from potassium is \( 4 \times +1 = +4 \).
- The total contribution from the cyanide ions is \( 6 \times -1 = -6 \).

Now, substituting these values into the equation:
\[ 4 + x - 6 = 0 \]

### Step 5: Solve for \( x \)
Rearranging the equation gives:
\[ x - 2 = 0 \]
So,
\[ x = +2 \]

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

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