In this reaction `4Al + 3O_(2) to 4Al^(3+) + 6O^(2-)` which of the following statement is incorrect ?

Options

A. It is a redox reaction
B. Metallic aluminium is a reducing agent
C. Metallic aluminium is oxidised to `Al^(3+)`
D. Metallic aluminium is reducing to `Al^(3+)`

Answer

To determine which statement is incorrect regarding the reaction \(4Al + 3O_2 \rightarrow 4Al^{3+} + 6O^{2-}\), we will analyze the oxidation states of the elements involved and the nature of the reaction.

### Step-by-Step Solution:

1. **Identify the Oxidation States**:
- In elemental form, the oxidation state of aluminum (Al) is 0.
- The oxidation state of oxygen (O) in \(O_2\) is also 0.

2. **Determine the Oxidation States in Products**:
- In the products, aluminum is present as \(Al^{3+}\), which has an oxidation state of +3.
- Oxygen in the form of \(O^{2-}\) has an oxidation state of -2.

3. **Identify Oxidation and Reduction**:
- **Oxidation**: Aluminum is oxidized as its oxidation state increases from 0 to +3. This means it loses electrons. The half-reaction can be written as:
\[ Al \rightarrow Al^{3+} + 3e^- \]
- **Reduction**: Oxygen is reduced as its oxidation state decreases from 0 to -2. This means it gains electrons. The half-reaction can be written as:
\[ O_2 + 4e^- \rightarrow 2O^{2-} \]

4. **Classify the Reaction**:
- Since both oxidation (of Al) and reduction (of O) occur, this reaction is classified as a redox reaction.

5. **Evaluate the Statements**:
- **Statement 1**: "It is a redox reaction." - This is correct.
- **Statement 2**: "Metallic aluminum is a reducing agent." - This is correct because it gets oxidized.
- **Statement 3**: "Metallic aluminum is oxidized to \(Al^{3+}\)." - This is correct as the oxidation state increases.
- **Statement 4**: "Metallic aluminum is reducing to \(Al^{3+}\)." - This is incorrect because aluminum is oxidized, not reduced.

### Conclusion:
The incorrect statement is **Statement 4**: "Metallic aluminum is reducing to \(Al^{3+}\)."

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