The state of hybridisation of `C_(2),C_(3),C_(5) and C_(6)` of the hydrocarbon -
`{:(" "CH_(3)" " CH_(3)),(" "|" "|),(CH_(3)-C-CH=CH-CH-C-=CH),(""^(7)" "^(6)| " "^(5)" "^(4)" "^(2) " "^(1)),(" "CH_(3)):}`

Options

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

Explanation

To determine the state of hybridization of the specified carbon atoms (C2, C3, C5, and C6) in the hydrocarbon structure provided, we will analyze the bonding around each carbon atom.

### Step-by-Step Solution:

1. **Identify the Structure**: The hydrocarbon given is CH₃-CH(CH₃)-C≡C-CH=CH-CH(CH₃). We need to focus on the specified carbon atoms: C2, C3, C5, and C6.

2. **Analyze C2**:
- C2 is involved in a triple bond (C≡C) with C1.
- A triple bond consists of one sigma bond and two pi bonds.
- Therefore, C2 has 2 sigma bonds (one with C1 and one with C3) and 2 pi bonds (from the triple bond).
- **Hybridization of C2**: Since it has 2 sigma and 2 pi bonds, C2 is **sp hybridized**.

3. **Analyze C3**:
- C3 is connected to C2 (triple bond), C4 (single bond), and two hydrogen atoms.
- C3 has 4 sigma bonds (one with C2, one with C4, and two with hydrogen).
- **Hybridization of C3**: Since it has 4 sigma bonds, C3 is **sp³ hybridized**.

4. **Analyze C5**:
- C5 is connected to C4 (single bond), C6 (double bond), and one hydrogen atom.
- C5 has 3 sigma bonds (one with C4, one with C6, and one with hydrogen) and 1 pi bond (from the double bond with C6).
- **Hybridization of C5**: Since it has 3 sigma bonds and 1 pi bond, C5 is **sp² hybridized**.

5. **Analyze C6**:
- C6 is connected to C5 (double bond), C7 (single bond), and two hydrogen atoms.
- C6 has 4 sigma bonds (one with C5, one with C7, and two with hydrogen).
- **Hybridization of C6**: Since it has 4 sigma bonds, C6 is **sp³ hybridized**.

### Summary of Hybridization:
- **C2**: sp
- **C3**: sp³
- **C5**: sp²
- **C6**: sp³