What is correct order of decreasing ionic character ?
`PbCl_2(I) , PbF_2(II), PbI_2(III), PbBr_2(IV)`

Options

A. III gt IV gt I gt II
B. I gt II gt III gt IV
C. III gt I gt IV gt II
D. II gt I gt IV gt III

Answer

To determine the correct order of decreasing ionic character for the compounds \( \text{PbCl}_2 \), \( \text{PbF}_2 \), \( \text{PbI}_2 \), and \( \text{PbBr}_2 \), we need to analyze the ionic and covalent character of these compounds based on the properties of the cations and anions involved.

### Step-by-Step Solution:

1. **Identify the Cation**:
All the compounds contain the same cation, \( \text{Pb}^{2+} \). Therefore, the cation's contribution to ionic character is constant across all compounds.

**Hint**: Focus on the anions since the cation is the same for all compounds.

2. **Identify the Anions**:
The anions in the compounds are:
- \( \text{Cl}^- \) (Chloride)
- \( \text{F}^- \) (Fluoride)
- \( \text{I}^- \) (Iodide)
- \( \text{Br}^- \) (Bromide)

3. **Analyze the Size of Anions**:
The size of the anions increases in the order:
\[ \text{F}^- < \text{Cl}^- < \text{Br}^- < \text{I}^- \]
This means that \( \text{F}^- \) is the smallest and \( \text{I}^- \) is the largest.

**Hint**: Remember that larger anions lead to increased covalent character due to their ability to polarize the cation.

4. **Apply Fajan's Rule**:
According to Fajan's rule, the ionic character decreases as the size of the anion increases. This is because larger anions can polarize the electron cloud of the cation more effectively, leading to a greater covalent character.

5. **Determine the Order of Ionic Character**:
Since \( \text{F}^- \) is the smallest anion, it will have the highest ionic character. Conversely, \( \text{I}^- \), being the largest, will have the lowest ionic character. Therefore, the order of decreasing ionic character is:
\[ \text{PbF}_2 > \text{PbCl}_2 > \text{PbBr}_2 > \text{PbI}_2 \]

6. **Final Order**:
The correct order of decreasing ionic character is:
\[ \text{PbF}_2 (II) > \text{PbCl}_2 (I) > \text{PbBr}_2 (IV) > \text{PbI}_2 (III) \]

### Conclusion:
Thus, the final answer is:
\[ \text{PbF}_2 > \text{PbCl}_2 > \text{PbBr}_2 > \text{PbI}_2 \]