The hybridization of `Fe` in `K_4[Fe(CN)_6]` complex is:

Options

A. (a) `d^2sp^2`
B. (b) `dsp^2`
C. (c) `d^2sp^3`
D. (d) `sp^3`

Explanation

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

### Step 1: Determine the oxidation state of iron (Fe)
1. The complex \( K_4[Fe(CN)_6] \) contains 4 potassium ions (K) and 6 cyanide ions (CN).
2. Each potassium ion has a charge of +1, contributing a total charge of +4.
3. Each cyanide ion has a charge of -1, contributing a total charge of -6.
4. The overall charge of the complex is neutral, so we can set up the equation:
\[ \text{Charge of Fe} + 4 (\text{from K}) + 6 (\text{from CN}) = 0 \]
\[ \text{Charge of Fe} + 4 - 6 = 0 \]
\[ \text{Charge of Fe} = +2 \]