In which of the following, do all the `C` atoms not have the same hybridisation ?

Options

A. `(CH_(3))_(3)C^(*)`
B. `CH_(3)-overset(ddot)overset(Θ)(CH)-CH_(3)`
C. `CH_(2)=C=CH_(2)`
D. `CH_(2)=overset(ddot)overset(Θ)(CH)`

Explanation

To determine in which of the given compounds all carbon atoms do not have the same hybridization, we need to analyze each compound's structure and the hybridization of its carbon atoms.

### Step-by-Step Solution:

1. **Identify the Compounds:**
Let's denote the compounds as:
- Compound A: \( \text{C}_3\text{H}_7 \) (C3 radical)
- Compound B: \( \text{CH}_3\text{CH}^- \text{CH}_3 \) (with a negative charge)
- Compound C: \( \text{CH}_2=\text{C}=\text{CH}_2 \) (1,2-propadiene)
- Compound D: \( \text{CH}_2=\text{C}=\text{CH}_2 \) (another structure with double bonds)

2. **Analyze Compound A:**
- Structure: \( \text{C}_3\text{H}_7 \) can be represented as \( \text{H}_3\text{C}-\text{C}^\cdot-\text{C} \).
- Hybridization: The terminal carbons (CH3) are \( \text{sp}^3 \) hybridized, and the central carbon (radical) is also \( \text{sp}^3 \) due to three surrounding atoms.
- Conclusion: All carbons have the same hybridization \( \text{sp}^3 \).

3. **Analyze Compound B:**
- Structure: \( \text{CH}_3\text{CH}^- \text{CH}_3 \).
- Hybridization: The two terminal carbons (CH3) are \( \text{sp}^3 \) hybridized, and the central carbon (with a negative charge) is also \( \text{sp}^3 \) due to three surrounding atoms.
- Conclusion: All carbons have the same hybridization \( \text{sp}^3 \).

4. **Analyze Compound C:**
- Structure: \( \text{CH}_2=\text{C}=\text{CH}_2 \).
- Hybridization: The terminal carbons (CH2) are \( \text{sp}^2 \) hybridized, and the central carbon (C) is \( \text{sp} \) hybridized due to two double bonds.
- Conclusion: Not all carbons have the same hybridization (sp, sp²).

5. **Analyze Compound D:**
- Structure: Similar to Compound C, \( \text{CH}_2=\text{C}=\text{CH}_2 \).
- Hybridization: Same as Compound C, the terminal carbons are \( \text{sp}^2 \) and the central carbon is \( \text{sp} \).
- Conclusion: Not all carbons have the same hybridization (sp, sp²).

### Final Answer:
The compounds in which not all carbon atoms have the same hybridization are **Compound C** and **Compound D**.