In the compound `HC-=C-CH=CH_2`, the hybridizations of `C-2` and `C-3` carbons are, respectively,

Options

A. `sp^3` and `sp^3`
B. `sp^2` and `sp^3`
C. `sp^2` and `sp`
D. `sp^3` and `sp`

Explanation

To determine the hybridization of the carbon atoms C2 and C3 in the compound \( HC \equiv C - C = CH_2 \), we will follow these steps:

### Step 1: Identify the Structure
The compound can be represented as follows:
- C1: \( H-C \equiv C \) (triple bond)
- C2: \( C \) (single bond to C3)
- C3: \( C \) (double bond to C4)
- C4: \( CH_2 \)

### Step 2: Number the Carbons
We will number the carbon atoms starting from the end of the chain with the double bond:
- C1: \( H-C \equiv C \) (1)
- C2: \( C \) (2)
- C3: \( C \) (3)
- C4: \( CH_2 \) (4)

### Step 3: Analyze C2 Hybridization
For C2:
- **Bonds**: C2 is bonded to one hydrogen atom and two other carbon atoms (C1 and C3).
- **Steric Number Calculation**:
- Number of bonded atoms = 3 (1 H + 2 C)
- Number of lone pairs = 0 (no lone pairs on carbon)
- Steric Number = 3 + 0 = 3
- **Hybridization**: A steric number of 3 corresponds to \( sp^2 \) hybridization.

### Step 4: Analyze C3 Hybridization
For C3:
- **Bonds**: C3 is bonded to one hydrogen atom and two other carbon atoms (C2 and C4).
- **Steric Number Calculation**:
- Number of bonded atoms = 2 (1 H + 1 C from C2 + 1 C from C4)
- Number of lone pairs = 0 (no lone pairs on carbon)
- Steric Number = 2 + 0 = 2
- **Hybridization**: A steric number of 2 corresponds to \( sp \) hybridization.

### Final Answer
The hybridizations of C2 and C3 are:
- C2: \( sp^2 \)
- C3: \( sp \)