The EAN of Ni in `[Ni(CN)_(4)]^(2-)` is

Options

A. 34
B. 35
C. 36
D. 28

Explanation

To find the Effective Atomic Number (EAN) of nickel in the complex \([Ni(CN)_{4}]^{2-}\), we can follow these steps:

### Step 1: Identify the atomic number of nickel (Ni)
The atomic number (Z) of nickel is 28.

### Step 2: Determine the oxidation state of nickel in the complex
The overall charge of the complex is -2. The cyanide ion (CN) has a charge of -1, and there are 4 cyanide ligands.

To find the oxidation state (OS) of nickel, we can set up the equation:
\[ \text{OS of Ni} + 4 \times (-1) = -2 \]
\[ \text{OS of Ni} - 4 = -2 \]
\[ \text{OS of Ni} = -2 + 4 = +2 \]

### Step 3: Calculate the EAN using the formula
The formula for EAN is:
\[ \text{EAN} = Z - \text{OS} + 2 \times \text{number of ligands} \]

Substituting the values we have:
- \(Z = 28\)
- \(\text{OS} = +2\)
- Number of ligands = 4

Now substituting these values into the formula:
\[ \text{EAN} = 28 - 2 + 2 \times 4 \]
\[ \text{EAN} = 28 - 2 + 8 \]
\[ \text{EAN} = 26 + 8 = 34 \]

### Final Answer
The EAN of nickel in \([Ni(CN)_{4}]^{2-}\) is **34**.

---