Only `sp and sp^2` hybrid orbitals are involved in the formation of
(a) `CH_3 - CH = CH_2`
(b) `CH_3 - CH_3`
(c ) `CH_3 - C -= CH_3`
(d) `CH_2 = C = CH_2`
(e ) `CH_3 - C -= C - CH_3`

Explanation

To determine which of the given molecules involves only `sp` and `sp^2` hybrid orbitals in their formation, we will analyze each compound based on the number of sigma bonds formed by the carbon atoms.

### Step-by-Step Solution:

1. **Analyze CH₃-CH=CH₂ (Propene)**:
- The first carbon (C1) in CH₃ has 4 sigma bonds (3 with H and 1 with C2) → **sp³ hybridized**.
- The second carbon (C2) has 3 sigma bonds (1 with C1 and 2 with C3) → **sp² hybridized**.
- The third carbon (C3) has 2 sigma bonds (1 with C2 and 1 with H) → **sp² hybridized**.
- Conclusion: Contains sp³ hybridized carbon.

2. **Analyze CH₃-CH₃ (Ethane)**:
- Both carbons have 4 sigma bonds (3 with H and 1 with the other C) → **sp³ hybridized**.
- Conclusion: Contains sp³ hybridized carbon.

3. **Analyze CH₃-C≡C-CH₃ (Butyne)**:
- The terminal carbons (C1 and C4) each have 2 sigma bonds (1 with H and 1 with C2 or C3) → **sp hybridized**.
- The internal carbons (C2 and C3) each have 1 sigma bond (with each other) and 2 pi bonds → **sp hybridized**.
- Conclusion: Contains only sp hybridized carbon.

4. **Analyze CH₂=C=CH₂ (Allene)**:
- The terminal carbons (C1 and C3) each have 2 sigma bonds (1 with H and 1 with C2) → **sp² hybridized**.
- The central carbon (C2) has 2 sigma bonds (with C1 and C3) → **sp hybridized**.
- Conclusion: Contains sp and sp² hybridized carbon.

5. **Analyze CH₃-C≡C-CH₃ (Butyne)**:
- The terminal carbons (C1 and C4) each have 2 sigma bonds (1 with H and 1 with C2 or C3) → **sp hybridized**.
- The internal carbons (C2 and C3) each have 1 sigma bond (with each other) and 2 pi bonds → **sp hybridized**.
- Conclusion: Contains only sp hybridized carbon.

### Final Conclusion:
Among the given compounds, only **CH₂=C=CH₂ (Allene)** contains carbons that are hybridized as `sp` and `sp²`. Therefore, the answer is:

**(d) CH₂=C=CH₂**