The hybrid orbitals that will form the compound `CH_3 - C -= C - CH_2 - CH_3` are:

Options

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

Answer

To determine the hybrid orbitals that will form the compound `CH3 - C ≡ C - CH2 - CH3`, we need to analyze the hybridization of each carbon atom in the compound step by step.

### Step 1: Identify the Structure of the Compound
The compound `CH3 - C ≡ C - CH2 - CH3` consists of two terminal carbon atoms (CH3 and CH2) and a central carbon-carbon triple bond (C ≡ C).

### Step 2: Determine the Hybridization of Each Carbon Atom
1. **Terminal Carbons (CH3 and CH2)**:
- The carbon in CH3 is bonded to three hydrogen atoms and one carbon atom. Therefore, it forms four sigma bonds.
- This carbon is **sp3 hybridized** because it uses one s orbital and three p orbitals to form four equivalent sp3 hybrid orbitals.

2. **Central Carbon (C ≡ C)**:
- The central carbon atom is involved in a triple bond with another carbon atom. A triple bond consists of one sigma bond and two pi bonds.
- This carbon is bonded to one carbon atom (forming a sigma bond) and has two remaining p orbitals that form the two pi bonds with the adjacent carbon.
- Therefore, this carbon is **sp hybridized** because it uses one s orbital and one p orbital to form two sp hybrid orbitals, which are used for the sigma bond, while the remaining two p orbitals are used for the pi bonds.

### Step 3: Summarize the Hybridization
- The hybridization of the terminal carbon (CH3) is **sp3**.
- The hybridization of the central carbon (C ≡ C) is **sp**.
- The other terminal carbon (CH2) is also **sp3** hybridized as it forms four sigma bonds (two with hydrogen and one with the adjacent carbon).

### Conclusion
The hybrid orbitals involved in the formation of the compound `CH3 - C ≡ C - CH2 - CH3` are **sp3 and sp**.