What is the hybridization of Fe in complex `K_3Fe(CN)_6` is ?

Options

A. `sp^3`
B. `dsp^3`
C. `sp^3d^2`
D. `d^2sp^3`

Answer

To determine the hybridization of iron (Fe) in the complex \( K_3Fe(CN)_6 \), we can follow these steps:

### Step 1: Determine the oxidation state of iron (Fe)
In the complex \( K_3Fe(CN)_6 \):
- Potassium (K) has a charge of +1. Since there are 3 potassium ions, their total contribution is \( +3 \).
- Each cyanide ion (CN) has a charge of -1. Since there are 6 cyanide ions, their total contribution is \( -6 \).

Let the oxidation state of iron be \( x \). The overall charge of the complex is neutral (0). Therefore, we can write the equation:

\[ x + 3 - 6 = 0 \]

Solving for \( x \):

\[ x - 3 = 0 \implies x = +3 \]

### Step 2: Write the electron configuration of iron
The atomic number of iron (Fe) is 26. The electron configuration of neutral iron is:

\[ [Ar] 4s^2 3d^6 \]

Since iron is in the +3 oxidation state, it loses 3 electrons. The electrons are removed first from the 4s subshell and then from the 3d subshell. Thus, the electron configuration for \( Fe^{3+} \) will be:

\[ [Ar] 3d^5 \]

### Step 3: Determine the geometry of the complex
The complex \( K_3Fe(CN)_6 \) contains 6 cyanide ligands, which are strong field ligands. The presence of 6 ligands indicates that the geometry of the complex is octahedral.

### Step 4: Determine the hybridization
In an octahedral complex, the hybridization can be determined by the number of ligands. For 6 ligands, the hybridization is \( d^2sp^3 \) or \( sp^3d^2 \).

In this case, since we have 5 electrons in the \( 3d \) subshell, we can pair up the electrons due to the strong field nature of the cyanide ligands. The pairing of electrons in the \( 3d \) orbitals leads to the formation of hybrid orbitals.

Thus, the hybridization of iron in \( K_3Fe(CN)_6 \) is:

\[ \text{Hybridization} = d^2sp^3 \]

### Final Answer
The hybridization of Fe in the complex \( K_3Fe(CN)_6 \) is \( d^2sp^3 \).

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