The conversion of sugar `C_(12)H_(22)O_(11) to CO_(2)` is :

Options

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

Explanation

To determine the type of reaction that the conversion of sugar \( C_{12}H_{22}O_{11} \) to \( CO_{2} \) represents, we can follow these steps:

### Step 1: Identify the Reactant and Product
The reactant is sugar, represented by the chemical formula \( C_{12}H_{22}O_{11} \), and the product is carbon dioxide, represented by \( CO_{2} \).

### Step 2: Define Oxidation and Reduction
- **Oxidation**: This is a process where the oxidation state of an element increases, which typically involves the loss of electrons.
- **Reduction**: This is the opposite process, where the oxidation state of an element decreases, typically involving the gain of electrons.

### Step 3: Determine the Oxidation State of Carbon in Sugar
To find the oxidation state of carbon in \( C_{12}H_{22}O_{11} \):
- Let the oxidation state of carbon be \( x \).
- The equation for the oxidation state can be set up as follows:
\[ 12x + 22(+1) + 11(-2) = 0 \]
Simplifying this gives:
\[ 12x + 22 - 22 = 0 \implies 12x = 0 \implies x = 0 \]
Thus, the oxidation state of carbon in \( C_{12}H_{22}O_{11} \) is 0.

### Step 4: Determine the Oxidation State of Carbon in Carbon Dioxide
Now, for \( CO_{2} \):
- Let the oxidation state of carbon be \( y \).
- The equation for the oxidation state can be set up as follows:
\[ y + 2(-2) = 0 \]
Simplifying this gives:
\[ y - 4 = 0 \implies y = +4 \]
Thus, the oxidation state of carbon in \( CO_{2} \) is +4.

### Step 5: Analyze the Change in Oxidation State
- The oxidation state of carbon changes from 0 in \( C_{12}H_{22}O_{11} \) to +4 in \( CO_{2} \).
- This increase in oxidation state indicates that carbon is losing electrons, which confirms that this process is an oxidation reaction.

### Conclusion
The conversion of sugar \( C_{12}H_{22}O_{11} \) to \( CO_{2} \) is classified as an **oxidation reaction**.

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