n the hydrocarbon,
`overset(6)CH_(3)-overset(5)CH=overset(4)CH-overset(3)CH_(2)-overset(2)C-=overset(1)CH`
The state of hybridization of carbons 1, 3 and 5 are in the following sequence :

Options

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

Explanation

To determine the state of hybridization of carbons 1, 3, and 5 in the given hydrocarbon structure, we will analyze each carbon atom based on the number of sigma bonds and lone pairs they have.

### Step-by-Step Solution:

1. **Identify the Structure**:
The structure provided is:
```
CH3-CH=CH-CH2-C≡CH
```
Here, we will label the carbon atoms from 1 to 6:
- C1: CH (attached to C2)
- C2: CH2 (attached to C3)
- C3: CH (attached to C4)
- C4: CH (attached to C5)
- C5: C≡C (triple bond with C6)
- C6: CH (attached to C5)

2. **Determine Hybridization of Carbon 1 (C1)**:
- C1 is connected to one hydrogen and one carbon (C2) through a single bond (sigma bond).
- It is involved in a triple bond with C6 (which consists of one sigma bond and two pi bonds).
- Total sigma bonds = 1 (C1 to C2) + 1 (C1 to C6) = 2.
- Since there are no lone pairs, the steric number is 2.
- Hybridization = sp (since steric number 2 corresponds to sp hybridization).

3. **Determine Hybridization of Carbon 3 (C3)**:
- C3 is connected to two hydrogens and two carbons (C2 and C4) through single bonds.
- Total sigma bonds = 2 (C3 to C2 and C3 to C4) + 2 (C3 to H) = 4.
- Since there are no lone pairs, the steric number is 4.
- Hybridization = sp³ (since steric number 4 corresponds to sp³ hybridization).

4. **Determine Hybridization of Carbon 5 (C5)**:
- C5 is connected to one carbon (C4) through a single bond and has a triple bond with C6.
- Total sigma bonds = 1 (C5 to C4) + 1 (C5 to C6) = 2.
- Since there are no lone pairs, the steric number is 3.
- Hybridization = sp² (since steric number 3 corresponds to sp² hybridization).

5. **Summarize the Hybridization**:
- C1: sp
- C3: sp³
- C5: sp²

### Final Answer:
The state of hybridization of carbons 1, 3, and 5 are in the sequence: **sp, sp³, sp²**.