If EAN of metal in `K_(4)[M(CN)_(6)]` is 36 find the number of d-electrons in complex.

Explanation

To solve the problem of finding the number of d-electrons in the complex \( K_4[M(CN)_6] \) given that the effective atomic number (EAN) of the metal is 36, we can follow these steps:

### Step 1: Write the formula for Effective Atomic Number (EAN)
The formula for EAN is given by:
\[ \text{EAN} = Z - \text{oxidation state} + 2 \times \text{coordination number} \]
Where:
- \( Z \) = atomic number of the metal
- Oxidation state = charge on the metal
- Coordination number = number of ligands attached to the metal

### Step 2: Identify the coordination number
In the complex \( K_4[M(CN)_6] \), the ligand is \( CN^- \) and there are 6 of them. Therefore, the coordination number is:
\[ \text{Coordination number} = 6 \]

### Step 3: Set up the equation for oxidation state
The overall charge of the complex \( K_4[M(CN)_6] \) is neutral. The \( K^+ \) ions contribute a charge of +4, and the \( CN^- \) ligands contribute a charge of -6. Thus, we can set up the equation for the oxidation state \( X \) of the metal \( M \):
\[ X - 6 = -4 \implies X = +2 \]

### Step 4: Substitute values into the EAN formula
Now we can substitute the values into the EAN formula:
\[ 36 = Z - 2 + 2 \times 6 \]
This simplifies to:
\[ 36 = Z - 2 + 12 \]
\[ 36 = Z + 10 \]
\[ Z = 36 - 10 = 26 \]

### Step 5: Identify the metal and its electron configuration
The atomic number \( Z = 26 \) corresponds to iron (Fe). The electron configuration of neutral iron is:
\[ \text{Fe: } [Ar] 3d^6 4s^2 \]
When iron is in the +2 oxidation state (as we found), it loses two electrons, typically from the 4s orbital:
\[ \text{Fe}^{2+}: [Ar] 3d^6 \]

### Step 6: Count the d-electrons
In the \( \text{Fe}^{2+} \) state, the number of d-electrons is:
\[ \text{Number of d-electrons} = 6 \]

### Final Answer
Thus, the number of d-electrons in the complex \( K_4[M(CN)_6] \) is **6**.

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