`PbCl_(4)` exists but `PbBr_(4)` and `PbI_(4)` do not because of

Explanation

In `Pbl_(4)`, the anion `I^(-)` is a very strong reducing agent and reduces `Pb^(4+)` ion to ` Pb^(2+)` ion. Thus, `Pbl_(4)` cannot be formed But in `PbCl_(4)`, anion `Cl^(-)` is not in a position to act as reducing agent. Tehrefore. It can exist.