In view of the signs of `Delta_(r)G^(@)` for the following reactions :
`PbO_(2)+Pbto2PbO,Delta_(r)G^(@)lt0`
`SnO_(2)+Snto2SnO,Delta_(r)G^(@)gt0`,
which oxidation states are more characteristic for lead and tin ?

Options

A. For lead + 2, for tin + 2
B. For lead + 4, for tin + 4
C. For lead + 2, for tin + 4
D. For lead + 4, for tin + 2

Answer

To determine the more characteristic oxidation states of lead (Pb) and tin (Sn) based on the given reactions and the signs of ΔG° (Gibbs free energy change), we can follow these steps:

### Step-by-Step Solution:

1. **Analyze the First Reaction:**
- The reaction is:
\[ \text{PbO}_2 + \text{Pb} \rightarrow 2\text{PbO} \]
- Given that ΔG° < 0, this indicates that the reaction is spontaneous.

2. **Determine the Oxidation States for Lead:**
- In PbO₂, the oxidation state of Pb can be calculated as follows:
\[ x - 4 = 0 \implies x = +4 \]
- In PbO, the oxidation state of Pb is:
\[ x - 2 = 0 \implies x = +2 \]
- Since the reaction is spontaneous (ΔG° < 0), the oxidation state of the product (PbO) is more characteristic. Therefore, the more characteristic oxidation state for lead is +2.

3. **Analyze the Second Reaction:**
- The reaction is:
\[ \text{SnO}_2 + \text{Sn} \rightarrow 2\text{SnO} \]
- Given that ΔG° > 0, this indicates that the reaction is non-spontaneous.

4. **Determine the Oxidation States for Tin:**
- In SnO₂, the oxidation state of Sn can be calculated as follows:
\[ x - 4 = 0 \implies x = +4 \]
- In SnO, the oxidation state of Sn is:
\[ x - 2 = 0 \implies x = +2 \]
- Since the reaction is non-spontaneous (ΔG° > 0), the oxidation state of the reactant (SnO₂) is more characteristic. Therefore, the more characteristic oxidation state for tin is +4.

5. **Conclusion:**
- The more characteristic oxidation state for lead (Pb) is +2, and for tin (Sn) is +4.

### Final Answer:
- Lead (Pb) has a more characteristic oxidation state of +2.
- Tin (Sn) has a more characteristic oxidation state of +4.