Oxidation number of carbon is zero in the compound

Options

A. methyl chloride
B. chloroform
C. glucose
D. carbon tetrachloride

Explanation

To determine the oxidation number of carbon in the given compounds, we will analyze each compound step by step. The compounds provided are methyl chloride, chloroform, glucose, and carbon tetrachloride.

### Step 1: Identify the molecular formulas
1. **Methyl Chloride (CH₃Cl)**
2. **Chloroform (CHCl₃)**
3. **Glucose (C₆H₁₂O₆)**
4. **Carbon Tetrachloride (CCl₄)**

### Step 2: Calculate the oxidation number of carbon in each compound

#### 1. Methyl Chloride (CH₃Cl)
- Let the oxidation state of carbon be \( x \).
- The oxidation state of hydrogen is +1 (3 hydrogen atoms) and chlorine is -1 (1 chlorine atom).
- The equation becomes:
\[ x + 3(+1) + 1(-1) = 0 \]
\[ x + 3 - 1 = 0 \implies x + 2 = 0 \implies x = -2 \]
- **Oxidation number of carbon in methyl chloride is -2.**

#### 2. Chloroform (CHCl₃)
- Let the oxidation state of carbon be \( x \).
- The oxidation state of hydrogen is +1 (1 hydrogen atom) and chlorine is -1 (3 chlorine atoms).
- The equation becomes:
\[ x + 1(+1) + 3(-1) = 0 \]
\[ x + 1 - 3 = 0 \implies x - 2 = 0 \implies x = +2 \]
- **Oxidation number of carbon in chloroform is +2.**

#### 3. Glucose (C₆H₁₂O₆)
- Let the oxidation state of carbon be \( x \) (6 carbon atoms).
- The oxidation state of hydrogen is +1 (12 hydrogen atoms) and oxygen is -2 (6 oxygen atoms).
- The equation becomes:
\[ 6x + 12(+1) + 6(-2) = 0 \]
\[ 6x + 12 - 12 = 0 \implies 6x = 0 \implies x = 0 \]
- **Oxidation number of carbon in glucose is 0.**

#### 4. Carbon Tetrachloride (CCl₄)
- Let the oxidation state of carbon be \( x \).
- The oxidation state of chlorine is -1 (4 chlorine atoms).
- The equation becomes:
\[ x + 4(-1) = 0 \]
\[ x - 4 = 0 \implies x = +4 \]
- **Oxidation number of carbon in carbon tetrachloride is +4.**

### Conclusion
The oxidation number of carbon is 0 in glucose. Therefore, the correct answer is glucose.