The number of electron in `t_(2g)` orbitals in `K_(4)[Fe(CN)_(6)]` is

Options

A. 2
B. 3
C. 6
D. 5

Explanation

To determine the number of electrons in the \( t_{2g} \) orbitals of the complex \( K_4[Fe(CN)_6] \), we will follow these steps:

### Step 1: Determine the oxidation state of iron (Fe)
In the complex \( K_4[Fe(CN)_6] \), potassium (K) has a +1 charge, and cyanide (CN) has a -1 charge.

- The overall charge of the complex is neutral, so:
\[ 4(+1) + x + 6(-1) = 0 \]
\[ 4 + x - 6 = 0 \]
\[ x - 2 = 0 \implies x = +2 \]

Thus, the oxidation state of iron in this complex is +2.

### Step 2: Determine the electron configuration of \( Fe^{2+} \)
The atomic number of iron (Fe) is 26. The electron configuration of neutral iron is:
\[ \text{Fe: } [Ar] 4s^2 3d^6 \]
When iron is in the +2 oxidation state, it loses two electrons from the 4s orbital:
\[ Fe^{2+}: [Ar] 3d^6 \]

### Step 3: Analyze the ligand field
Cyanide (CN) is a strong field ligand. In the presence of strong field ligands, the \( d \)-orbitals split into two groups: \( t_{2g} \) and \( e_g \). The \( t_{2g} \) orbitals can hold a maximum of 6 electrons, while the \( e_g \) orbitals can hold a maximum of 4 electrons.

### Step 4: Fill the \( d \)-orbitals according to Hund's rule and the Pauli exclusion principle
For \( Fe^{2+} \) with a \( 3d^6 \) configuration, the electrons will fill the \( t_{2g} \) orbitals first due to the strong field nature of CN, which causes pairing of electrons in the \( t_{2g} \) orbitals:

1. The first three electrons will occupy the three \( t_{2g} \) orbitals singly (following Hund's rule).
2. The next three electrons will pair up in the \( t_{2g} \) orbitals.

Thus, the distribution will be:
- \( t_{2g} \): 6 electrons (3 pairs)
- \( e_g \): 0 electrons

### Conclusion
The number of electrons in the \( t_{2g} \) orbitals of \( K_4[Fe(CN)_6] \) is **6**.

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