Select the response that correctly identifies the number of carbon atoms of each type of hybridization in the compound given below
`H_(2)=C=CH-CH=O`

Options

A. `{:(sp^(3),sp^(2),sp),(2,2,0):}`
B. `{:(sp^(3),sp^(2),sp),(1,3,0):}`
C. `{:(sp^(3),sp^(2),sp),(0,3,1):}`
D. `{:(sp^(3),sp^(2),sp),(1,2,1):}`

Explanation

To determine the number of carbon atoms of each type of hybridization in the compound \( H_2C=CH-CH=O \), we will analyze the structure of the compound step by step.

### Step 1: Write the structure of the compound
The compound can be expanded as follows:
- The first carbon (C1) is bonded to two hydrogens and has a double bond with the second carbon (C2).
- The second carbon (C2) has a double bond with the first carbon (C1) and a single bond with the third carbon (C3).
- The third carbon (C3) has a double bond with the oxygen (O) and a single bond with the second carbon (C2).

The structure can be represented as:
```
H H
\ /
C1
||
C2 - C3 = O
```

### Step 2: Identify the hybridization of each carbon atom
1. **C1**:
- C1 has 2 single bonds (to H) and 1 double bond (to C2).
- This means it has 3 sigma bonds and 1 pi bond.
- Hybridization = **SP2** (3 sigma + 1 pi).

2. **C2**:
- C2 has 1 double bond (to C1) and 1 single bond (to C3).
- This means it has 3 sigma bonds and 1 pi bond.
- Hybridization = **SP2** (3 sigma + 1 pi).

3. **C3**:
- C3 has 1 double bond (to O) and 1 single bond (to C2).
- This means it has 2 sigma bonds and 2 pi bonds.
- Hybridization = **SP** (2 sigma + 2 pi).

### Step 3: Count the number of carbon atoms for each type of hybridization
- **SP3**: 0 carbon atoms (none of the carbons have 4 sigma bonds).
- **SP2**: 2 carbon atoms (C1 and C2).
- **SP**: 1 carbon atom (C3).

### Final Count
- SP3: 0
- SP2: 2
- SP: 1

### Conclusion
The correct identification of the number of carbon atoms of each type of hybridization in the compound \( H_2C=CH-CH=O \) is:
- SP3: 0
- SP2: 2
- SP: 1