In `K_(4)[Fe(CN)_(6)],Fe` is in the form of

Options

A. An atom
B. Neutral complex
C. Cationic complex
D. Anionic complex

Answer

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

### Step 1: Identify the components of the complex
The complex \( K_4[Fe(CN)_6] \) consists of potassium ions (K) and the anionic complex ion \( [Fe(CN)_6]^{4-} \).

### Step 2: Determine the charge of the complex ion
Since there are 4 potassium ions (K) each with a +1 charge, the total positive charge contributed by potassium is \( 4 \times (+1) = +4 \).

### Step 3: Set up the charge balance equation
The overall charge of the compound must be neutral. Therefore, the charge of the anionic complex \( [Fe(CN)_6] \) must balance out the +4 charge from the potassium ions. This means:

\[ \text{Charge of } [Fe(CN)_6] + 4 = 0 \]

Thus, the charge of \( [Fe(CN)_6] \) must be -4.

### Step 4: Analyze the ligands
The cyanide ion (CN) is a monodentate ligand with a -1 charge. Since there are 6 cyanide ions in the complex, the total negative charge contributed by the cyanide ions is:

\[ 6 \times (-1) = -6 \]

### Step 5: Calculate the oxidation state of iron
Let the oxidation state of iron (Fe) be \( x \). The total charge of the complex can be expressed as:

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

Solving for \( x \):

\[ x - 6 = -4 \\ x = -4 + 6 \\ x = +2 \]

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

### Final Answer
Fe is in the form of +2 oxidation state.

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