The EAN of platinum in potassium hexachloridoplatinate (IV) is (Atomic number of Pt=78)

Options

A. 90
B. 86
C. 76
D. 88

Answer

To find the Effective Atomic Number (EAN) of platinum in potassium hexachloridoplatinate (IV), we will follow these steps:

### Step 1: Identify the oxidation state of platinum in the complex
The complex is potassium hexachloridoplatinate (IV), which indicates that platinum (Pt) is in the +4 oxidation state.

### Step 2: Determine the charge of the complex
The complex ion is [PtCl6]^(2-). Since each chlorine (Cl) has a -1 charge and there are 6 chlorines, the total negative charge contributed by chlorine is:
\[ 6 \times (-1) = -6 \]
Let the charge of the platinum ion be \( x \). The overall charge of the complex ion is:
\[ x + (-6) = -2 \]
Solving for \( x \):
\[ x - 6 = -2 \]
\[ x = -2 + 6 \]
\[ x = +4 \]
This confirms that platinum is in the +4 oxidation state.

### Step 3: Write the formula for the compound
Since potassium (K) has a +1 charge, and there are two potassium ions to balance the -2 charge of the complex ion, the formula for the compound is:
\[ K_2[PtCl_6] \]

### Step 4: Calculate the Effective Atomic Number (EAN)
The formula for EAN is:
\[ \text{EAN} = Z + 2n - x \]
Where:
- \( Z \) = Atomic number of the metal (platinum = 78)
- \( n \) = Number of ligands (6 chlorines)
- \( x \) = Oxidation state of the metal (+4)

Now substituting the values:
\[ \text{EAN} = 78 + 2(6) - 4 \]

### Step 5: Perform the calculations
Calculating the values:
1. \( 2 \times 6 = 12 \)
2. Now substitute back:
\[ \text{EAN} = 78 + 12 - 4 \]
\[ \text{EAN} = 78 + 12 = 90 \]
\[ \text{EAN} = 90 - 4 = 86 \]

### Final Answer
The Effective Atomic Number (EAN) of platinum in potassium hexachloridoplatinate (IV) is **86**.

---