Which of the following organic compounds has same hybridization as its combustion product `(CO_(2))`?

Options

A. Ethanol
B. Ethyne
C. Ethene
D. Ethane

Answer

To determine which organic compound has the same hybridization as its combustion product \( CO_2 \), we need to analyze the hybridization of \( CO_2 \) and compare it with the hybridization of the given organic compounds.

### Step-by-Step Solution:

1. **Identify the Hybridization of \( CO_2 \)**:
- The central atom in \( CO_2 \) is carbon (C).
- Carbon has 4 valence electrons.
- The formula for calculating hybridization is:
\[ \text{Hybridization} = \frac{1}{2} \left( \text{Number of valence electrons on central atom} + \text{Number of monovalent atoms} + \text{Charge} \right) \]
- For \( CO_2 \):
- Valence electrons on carbon = 4
- Number of monovalent atoms = 0 (no hydrogen or other monovalent atoms directly bonded to carbon)
- Charge = 0
- Plugging in the values:
\[ \text{Hybridization} = \frac{1}{2} (4 + 0 + 0) = \frac{4}{2} = 2 \]
- A hybridization value of 2 corresponds to \( sp \) hybridization.

2. **Analyze the Given Organic Compounds**:
- We need to find which of the following compounds has \( sp \) hybridization:
1. Ethanol (\( C_2H_5OH \))
2. Ethyne (\( C_2H_2 \))
3. Ethene (\( C_2H_4 \))
4. Ethane (\( C_2H_6 \))

3. **Determine the Hybridization of Each Compound**:
- **Ethanol (\( C_2H_5OH \))**:
- Structure: \( C_2H_5OH \) has all single bonds.
- Hybridization: Each carbon is \( sp^3 \).

- **Ethyne (\( C_2H_2 \))**:
- Structure: \( C \equiv C \) (triple bond between two carbons).
- Hybridization: Each carbon is \( sp \) (one triple bond).

- **Ethene (\( C_2H_4 \))**:
- Structure: \( C=C \) (double bond between two carbons).
- Hybridization: Each carbon is \( sp^2 \).

- **Ethane (\( C_2H_6 \))**:
- Structure: All single bonds.
- Hybridization: Each carbon is \( sp^3 \).

4. **Conclusion**:
- The only compound that has \( sp \) hybridization, like \( CO_2 \), is **Ethyne (\( C_2H_2 \))**.

### Final Answer:
Ethyne (\( C_2H_2 \)) has the same hybridization as its combustion product \( CO_2 \).