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
In the complex \( K_3Fe(CN)_6 \), potassium (K) has a +1 oxidation state. The cyanide ion (CN) has a -1 oxidation state.

The overall charge of the complex is neutral (0). Therefore, we can set up the equation:

\[ 3(+1) + x + 6(-1) = 0 \]

Where \( x \) is the oxidation state of iron.

Solving this gives:

\[ 3 - 6 + x = 0 \\ x = +3 \]

Thus, the oxidation state of iron in this complex is +3.

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

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

Since iron is in the +3 oxidation state, it loses three electrons. The electrons are lost first from the 4s orbital and then from the 3d orbital. Therefore, the electronic configuration of \( Fe^{3+} \) is:

\[ 3d^5 \]

### Step 3: Determine the nature of the ligands
Cyanide (CN) is a strong field ligand. Strong field ligands cause pairing of electrons in the d-orbitals, leading to low-spin complexes.

### Step 4: Pair the electrons in the d-orbitals
In the \( 3d \) subshell, the five electrons will pair up due to the influence of the strong field ligand. The configuration after pairing will look like this:

- 3d: ↑↓ ↑↓ ↑ ↑ (2 paired, 3 unpaired)

### Step 5: Determine the hybridization
In the case of \( Fe^{3+} \) with a \( 3d^5 \) configuration and the presence of 6 cyanide ligands, we need to consider the orbitals involved in hybridization.

The hybridization involves:
- 2 d-orbitals
- 1 s-orbital
- 3 p-orbitals

Thus, the hybridization is \( d^2sp^3 \).

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

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