The oxidation number of C in glucose `(C_(6)H_(12)O_(6))` is

Options

A. 0
B. `+22`
C. `+6`
D. `-6`

Explanation

To determine the oxidation number of carbon in glucose (C₆H₁₂O₆), we can follow these steps:

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

### Step 2: Set up the equation
- Let the oxidation number of carbon (C) be represented as \( x \).
- Since there are 6 carbon atoms in glucose, the contribution from carbon will be \( 6x \).
- The total contribution from hydrogen will be \( 12 \times (+1) = +12 \).
- The total contribution from oxygen will be \( 6 \times (-2) = -12 \).

### Step 3: Write the overall charge balance equation
- The overall charge of the molecule is neutral (0), so we can write the equation:
\[ 6x + 12 - 12 = 0 \]

### Step 4: Simplify the equation
- This simplifies to:
\[ 6x = 0 \]

### Step 5: Solve for \( x \)
- Dividing both sides by 6 gives:
\[ x = 0 \]

### Conclusion
- The oxidation number of carbon in glucose (C₆H₁₂O₆) is **0**.

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