`PbF_(4), PbCl_(4)` exist but `PbBr_(4)` do not exist because of

Options

A. large size of `Br^(-) and I^(-)`
B. strong oxidising character of `Pb^(4+)`
C. strong reducing character of `Pb^(4+)`
D. low electronegativity of `Br^(-) and I^(-)`

Explanation

To understand why PbBr₄ does not exist while PbF₄ and PbCl₄ do, we need to analyze the properties of the halogens involved and their interactions with lead (Pb) in different oxidation states.

### Step-by-Step Solution:

1. **Identify the Compounds**: We are considering the compounds PbF₄, PbCl₄, and PbBr₄. Lead can exist in multiple oxidation states, but here we focus on the +4 oxidation state.

2. **Oxidation States of Lead**: Lead can exist in +2 and +4 oxidation states. The formation of PbF₄ and PbCl₄ indicates that these halogens can oxidize lead from +2 to +4.

3. **Properties of Halogens**:
- Fluorine (F) and chlorine (Cl) are strong oxidizing agents. They have high electronegativities and can effectively oxidize lead from +2 to +4.
- Bromine (Br) and iodine (I), on the other hand, are weaker oxidizing agents. They have lower electronegativities compared to fluorine and chlorine.

4. **Electronegativity Trend**: As we move down the group of halogens (from fluorine to iodine), the electronegativity decreases. This means that bromine and iodine are less capable of attracting electrons and oxidizing other elements.

5. **Formation of PbBr₄**: Since bromine is a weak oxidizing agent, it cannot oxidize lead from +2 to +4. Therefore, PbBr₄ cannot form. In contrast, fluorine and chlorine can oxidize lead effectively, resulting in the formation of PbF₄ and PbCl₄.

6. **Conclusion**: The non-existence of PbBr₄ is due to the weak oxidizing character of bromine, which is unable to oxidize lead to the +4 oxidation state. Thus, while PbF₄ and PbCl₄ exist, PbBr₄ does not.

### Final Answer:
PbBr₄ does not exist because bromine is a weak oxidizing agent and cannot oxidize lead from +2 to +4.

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