The oxidation number of carbon in `C_(12) H_(22) O_(11)` is

Options

A. 0
B. `-6`
C. `+6`
D. `+2`

Answer

To find the oxidation number of carbon in the compound \( C_{12}H_{22}O_{11} \) (sucrose), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Compound**: We are given the molecular formula \( C_{12}H_{22}O_{11} \).

2. **Assign Oxidation Numbers**:
- The oxidation number of hydrogen (H) is generally +1.
- The oxidation number of oxygen (O) is generally -2.

3. **Set Up the Equation**:
- Let the oxidation number of carbon (C) be \( x \).
- There are 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms in the molecule.
- The total charge of the molecule is 0 (neutral compound).
- Therefore, we can set up the equation based on the oxidation numbers:
\[ 12x + 22(+1) + 11(-2) = 0 \]

4. **Simplify the Equation**:
- Substitute the oxidation numbers into the equation:
\[ 12x + 22 - 22 = 0 \]
- This simplifies to:
\[ 12x = 0 \]

5. **Solve for \( x \)**:
- Divide both sides by 12:
\[ x = \frac{0}{12} = 0 \]

6. **Conclusion**:
- The oxidation number of carbon in \( C_{12}H_{22}O_{11} \) is 0.