`PbI_(4)` does not exist because:

Options

A. iodine is not a reactive
B. `Pb(IV)` is oxidizing and `I^(-)` is storng reducing agent
C. Pb(IV) is less table than Pb(II)
D. `Pb^(4+)` is not easily formed

Answer

To determine why \( \text{PbI}_4 \) does not exist, let's analyze the situation step by step:

### Step 1: Understand the oxidation states
Lead (Pb) can exist in multiple oxidation states, primarily +2 and +4. In the case of \( \text{PbI}_4 \), lead would be in the +4 oxidation state.

**Hint:** Remember that elements can have multiple oxidation states, and the stability of these states can vary.

### Step 2: Analyze the reactivity of iodine
Iodine can exist in various oxidation states, including -1 (as iodide, \( \text{I}^- \)), 0 (as \( \text{I}_2 \)), +1, +3, +5, and +7. The iodide ion (\( \text{I}^- \)) is a strong reducing agent, meaning it can donate electrons easily.

**Hint:** Consider the reactivity of halogens and their ability to form compounds in different oxidation states.

### Step 3: Examine the properties of \( \text{Pb}^{4+} \)
The \( \text{Pb}^{4+} \) ion is a strong oxidizing agent. This means it can accept electrons from other species, such as \( \text{I}^- \).

**Hint:** Identify the roles of oxidizing and reducing agents in chemical reactions.

### Step 4: Reaction analysis
When \( \text{Pb}^{4+} \) reacts with \( \text{I}^- \), the \( \text{I}^- \) can reduce \( \text{Pb}^{4+} \) to \( \text{Pb}^{2+} \) while itself being oxidized to \( \text{I}_2 \). This reaction indicates that \( \text{I}^- \) is not a strong enough reducing agent to stabilize \( \text{Pb}^{4+} \) in the form of \( \text{PbI}_4 \).

**Hint:** Consider the stability of the products formed in redox reactions.

### Step 5: Conclusion on the existence of \( \text{PbI}_4 \)
Since \( \text{Pb}^{4+} \) is oxidizing and \( \text{I}^- \) is a strong reducing agent, the formation of \( \text{PbI}_4 \) is not feasible. Instead, the more stable compound \( \text{PbI}_2 \) is formed, which corresponds to the \( \text{Pb}^{2+} \) state.

**Hint:** Reflect on how the stability of oxidation states affects the formation of compounds.

### Final Answer
Thus, \( \text{PbI}_4 \) does not exist primarily because \( \text{Pb}^{4+} \) is a strong oxidizing agent, and \( \text{I}^- \) is a strong reducing agent, leading to the formation of \( \text{PbI}_2 \) instead.