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

Options

A. `+4`
B. `+3`
C. `+2`
D. 0

Answer

To find the oxidation number of carbon (C) in sucrose (C₁₂H₂₂O₁₁), we can follow these steps:

### Step 1: Assign oxidation numbers to known elements
- Hydrogen (H) typically has an oxidation number of +1.
- Oxygen (O) typically has an oxidation number of -2.

### Step 2: Set up the equation
Let the oxidation number of carbon be \( x \). In sucrose, we have:
- 12 carbon atoms
- 22 hydrogen atoms
- 11 oxygen atoms

The overall charge of the molecule is neutral (0).

### Step 3: Write the equation based on the oxidation numbers
The equation can be set up as follows:
\[ 12x + (22 \times 1) + (11 \times -2) = 0 \]
This simplifies to:
\[ 12x + 22 - 22 = 0 \]

### Step 4: Simplify the equation
The equation simplifies to:
\[ 12x = 0 \]

### Step 5: Solve for \( x \)
Dividing both sides by 12 gives:
\[ x = 0 \]

### Conclusion
The oxidation number of carbon in sucrose (C₁₂H₂₂O₁₁) is **0**.