Which of the following compounds contain all the carbon atoms in the same hybridisation state ?

Options

A. `H - C -= C - -= C - H`
B. `CH_(3) - C -= C - CH_(3)`
C. `CH_(2) = C = CH_(2)`
D. `CH_(2) = CH - CH = CH_(2)`

Explanation

To determine which of the given compounds contains all carbon atoms in the same hybridization state, we need to analyze the hybridization of carbon atoms in each compound. The hybridization can be determined by counting the number of sigma bonds around each carbon atom.

### Step-by-Step Solution:

1. **Understand Hybridization**:
- Carbon can undergo hybridization to form different types of bonds:
- **sp3**: 4 sigma bonds (single bonds)
- **sp2**: 3 sigma bonds (one double bond)
- **sp**: 2 sigma bonds (one triple bond)

2. **Analyze Each Compound**:
- **Option A: CH≡C-CH≡CH**
- Each carbon in this compound is involved in a triple bond with one sigma bond and two pi bonds.
- Each carbon has **2 sigma bonds**, so the hybridization for each carbon is **sp**.
- **Conclusion**: All carbons are **sp** hybridized.

- **Option B: CH3-C≡C-CH3**
- The terminal carbons (CH3) have **4 sigma bonds** (3 from H and 1 from C), so they are **sp3** hybridized.
- The central carbon (C≡C) has **2 sigma bonds**, so it is **sp** hybridized.
- **Conclusion**: Not all carbons are the same hybridization.

- **Option C: CH2=CH-CH2**
- The first carbon (CH2) has **2 sigma bonds** (1 from H and 1 from C) and **1 pi bond**, so it is **sp2** hybridized.
- The second carbon (CH) has **3 sigma bonds** (1 from H and 2 from C), so it is also **sp2** hybridized.
- The third carbon (CH2) has **2 sigma bonds** (1 from H and 1 from C) and **1 pi bond**, so it is **sp2** hybridized.
- **Conclusion**: All carbons are **sp2** hybridized.

- **Option D: CH2=CH-CH=CH2**
- The first carbon (CH2) has **2 sigma bonds** and **1 pi bond**, so it is **sp2** hybridized.
- The second carbon (CH) has **3 sigma bonds** (1 from H and 2 from C), so it is **sp2** hybridized.
- The third carbon (CH) has **3 sigma bonds** (1 from H and 2 from C), so it is **sp2** hybridized.
- The fourth carbon (CH2) has **2 sigma bonds** and **1 pi bond**, so it is **sp2** hybridized.
- **Conclusion**: All carbons are **sp2** hybridized.

3. **Final Conclusion**:
- The compounds that contain all carbon atoms in the same hybridization state are **Option A (sp)** and **Option D (sp2)**.

### Summary of Answers:
- **Correct Options**: A and D.