The conversion of `PbO_(2)` to `Pb(NO_(3))_(2)` is

Options

A. Oxidation
B. Reduction
C. Neither oxidation nor reduction
D. Both oxidation and reaction

Explanation

To determine whether the conversion of \( \text{PbO}_2 \) to \( \text{Pb(NO}_3)_2 \) involves oxidation, reduction, or neither, we need to analyze the oxidation states of lead (Pb) in both compounds.

**Step 1: Identify the oxidation state of Pb in \( \text{PbO}_2 \)**

In \( \text{PbO}_2 \):
- Oxygen (O) typically has an oxidation state of -2.
- Since there are two oxygen atoms, the total contribution from oxygen is \( 2 \times (-2) = -4 \).
- Let the oxidation state of Pb be \( x \). The overall charge of the compound is neutral (0), so we can set up the equation:

\[ x + (-4) = 0 \]

Solving for \( x \):

\[ x = +4 \]

Thus, the oxidation state of Pb in \( \text{PbO}_2 \) is +4.

**Step 2: Identify the oxidation state of Pb in \( \text{Pb(NO}_3)_2 \)**

In \( \text{Pb(NO}_3)_2 \):
- Each nitrate ion (\( \text{NO}_3^- \)) has an oxidation state of -1. Since there are two nitrate ions, the total contribution from the nitrate ions is \( 2 \times (-1) = -2 \).
- Let the oxidation state of Pb be \( y \). The overall charge of the compound is also neutral (0), so we can set up the equation:

\[ y + (-2) = 0 \]

Solving for \( y \):

\[ y = +2 \]

Thus, the oxidation state of Pb in \( \text{Pb(NO}_3)_2 \) is +2.

**Step 3: Compare the oxidation states of Pb in both compounds**

- In \( \text{PbO}_2 \), Pb has an oxidation state of +4.
- In \( \text{Pb(NO}_3)_2 \), Pb has an oxidation state of +2.

**Step 4: Determine if oxidation or reduction occurs**

- The oxidation state of Pb decreases from +4 to +2. A decrease in oxidation state indicates that reduction has occurred.

**Conclusion:**

The conversion of \( \text{PbO}_2 \) to \( \text{Pb(NO}_3)_2 \) is a reduction process.

**Final Answer: Reduction**

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