In which of the following compounds the oxidation number of carbon is maximum

Options

A. `HCHO`
B. `CHCl_(3)`
C. `CH_(3)OH`
D. `C_(12)H_(22)O_(11)`

Explanation

To determine the compound in which the oxidation number of carbon is maximum, we will analyze the oxidation state of carbon in each of the given compounds step by step.

### Step 1: Analyze HCHO (Formaldehyde)
1. Let the oxidation number of carbon be \( x \).
2. The oxidation number of hydrogen (H) is \( +1 \) and that of oxygen (O) is \( -2 \).
3. The compound has 2 hydrogen atoms and 1 oxygen atom.
4. The equation for the overall charge (which is 0) is:
\[ 2(+1) + x + 1(-2) = 0 \]
Simplifying this:
\[ 2 + x - 2 = 0 \implies x = 0 \]
5. **Oxidation number of carbon in HCHO is 0.**

### Step 2: Analyze CHCl₃ (Chloroform)
1. Let the oxidation number of carbon be \( x \).
2. The oxidation number of hydrogen is \( +1 \) and chlorine (Cl) is \( -1 \).
3. There are 3 chlorine atoms in the compound.
4. The equation for the overall charge is:
\[ 1(+1) + x + 3(-1) = 0 \]
Simplifying this:
\[ 1 + x - 3 = 0 \implies x - 2 = 0 \implies x = +2 \]
5. **Oxidation number of carbon in CHCl₃ is +2.**

### Step 3: Analyze CH₃OH (Methanol)
1. Let the oxidation number of carbon be \( x \).
2. The oxidation number of hydrogen is \( +1 \) and oxygen is \( -2 \).
3. There are 3 hydrogen atoms in the compound.
4. The equation for the overall charge is:
\[ 3(+1) + x + (-2) = 0 \]
Simplifying this:
\[ 3 + x - 2 = 0 \implies x + 1 = 0 \implies x = -1 \]
5. **Oxidation number of carbon in CH₃OH is -1.**

### Step 4: Analyze C₁₂H₂₂O₁₁ (Sucrose)
1. Let the oxidation number of carbon be \( x \).
2. The oxidation number of hydrogen is \( +1 \) and oxygen is \( -2 \).
3. There are 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms.
4. The equation for the overall charge is:
\[ 12x + 22 + 11(-2) = 0 \]
Simplifying this:
\[ 12x + 22 - 22 = 0 \implies 12x = 0 \implies x = 0 \]
5. **Oxidation number of carbon in C₁₂H₂₂O₁₁ is 0.**

### Conclusion
Now, we compare the oxidation numbers of carbon in all four compounds:
- HCHO: 0
- CHCl₃: +2
- CH₃OH: -1
- C₁₂H₂₂O₁₁: 0

The maximum oxidation number of carbon is **+2** in **CHCl₃ (Chloroform)**.

### Final Answer
**The compound with the maximum oxidation number of carbon is CHCl₃.**

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