In `overset(1)CH_(2)=overset(2)CH-overset(3)CH_(2)-overset(4)C-=overset(5)CH`, the `C_(2)-C_(3)` bond involves the hybridization of the type :

Options

A. `sp-sp^(2)`
B. `sp^(3)-sp^(3)`
C. `sp-sp^(3)`
D. `sp^(2)-sp^(3)`

Explanation

To determine the hybridization of the bond between carbon atoms C2 and C3 in the given compound `CH2=CH-CH2-C≡CH`, we can follow these steps:

### Step 1: Identify the Structure
The compound can be represented as:
```
H H
\ /
C2
||
C1
|
C3
/ \
H C4
/ \
H C5
```
Where:
- C1 is the carbon involved in a double bond with C2.
- C2 is the carbon we are analyzing.
- C3 is the next carbon in the chain.

### Step 2: Count the Sigma Bonds and Lone Pairs
To determine the hybridization, we need to calculate the steric number using the formula:
\[ \text{Steric Number} = \text{Number of Sigma Bonds} + \text{Number of Lone Pairs} \]

#### For C2:
- C2 has:
- 1 sigma bond with C1 (double bond contributes 1 sigma)
- 1 sigma bond with C3
- 2 sigma bonds with the two hydrogen atoms

Thus, the total number of sigma bonds for C2 is:
\[ 1 + 1 + 2 = 4 \]
Since there are no lone pairs on C2, the steric number for C2 is 3.

#### For C3:
- C3 has:
- 1 sigma bond with C2
- 1 sigma bond with C4
- 2 sigma bonds with the two hydrogen atoms

Thus, the total number of sigma bonds for C3 is:
\[ 1 + 1 + 2 = 4 \]
Since there are no lone pairs on C3, the steric number for C3 is also 4.

### Step 3: Determine the Hybridization
Using the steric number:
- If the steric number is 2, the hybridization is sp.
- If the steric number is 3, the hybridization is sp².
- If the steric number is 4, the hybridization is sp³.

From our calculations:
- C2 has a steric number of 3, so it is **sp² hybridized**.
- C3 has a steric number of 4, so it is **sp³ hybridized**.

### Conclusion
The bond between C2 and C3 involves the hybridization of the type **sp² for C2 and sp³ for C3**.

### Final Answer
The C2-C3 bond involves the hybridization of the type **sp² and sp³**.

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