The EAN of platinum in potassium hexachloroplatinate (IV) is:

Options

A. 46
B. 86
C. 36
D. 84

Answer

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

### Step 1: Identify the central metal atom and its oxidation state
The central metal atom in potassium hexachloroplatinate (IV) is platinum (Pt). The oxidation state of platinum in this complex is +4.

### Step 2: Determine the atomic number of platinum
The atomic number of platinum (Pt) is 78.

### Step 3: Calculate the number of electrons donated by the ligands
In this complex, there are 6 chloride ions (Cl⁻) as ligands. Each chloride ion donates 2 electrons. Therefore, the total number of electrons donated by the ligands can be calculated as follows:
- Number of chloride ligands = 6
- Electrons donated by each chloride = 2
- Total electrons donated = 6 × 2 = 12 electrons

### Step 4: Apply the EAN formula
The formula for the Effective Atomic Number (EAN) is given by:
\[ \text{EAN} = Z - n + e \]
Where:
- \( Z \) = Atomic number of the central metal atom
- \( n \) = Oxidation state of the central metal atom
- \( e \) = Total number of electrons donated by the ligands

Substituting the values:
- \( Z = 78 \) (atomic number of platinum)
- \( n = 4 \) (oxidation state of platinum)
- \( e = 12 \) (total electrons donated by ligands)

Now, substituting these values into the formula:
\[ \text{EAN} = 78 - 4 + 12 \]
\[ \text{EAN} = 78 - 4 + 12 = 86 \]

### Conclusion
The Effective Atomic Number (EAN) of platinum in potassium hexachloroplatinate (IV) is 86.