The hybridization of carbon in diamond, graphite and acetylene are respectively

Options

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

Explanation

To determine the hybridization of carbon in diamond, graphite, and acetylene, we can analyze the bonding and geometry of each structure step by step.

### Step-by-Step Solution:

1. **Hybridization in Diamond:**
- Diamond has a tetrahedral structure. Each carbon atom in diamond is bonded to four other carbon atoms.
- The hybridization involved is **sp³** because one s orbital and three p orbitals combine to form four equivalent sp³ hybrid orbitals.
- Therefore, the hybridization of carbon in diamond is **sp³**.

2. **Hybridization in Graphite:**
- Graphite has a planar structure where each carbon atom is bonded to three other carbon atoms, forming a hexagonal lattice.
- The hybridization in graphite is **sp²** because one s orbital and two p orbitals hybridize to form three sp² hybrid orbitals, leaving one unhybridized p orbital for π bonding.
- Therefore, the hybridization of carbon in graphite is **sp²**.

3. **Hybridization in Acetylene:**
- Acetylene (C₂H₂) has a linear structure. Each carbon atom is bonded to one hydrogen atom and to the other carbon atom via a triple bond.
- The hybridization in acetylene is **sp** because one s orbital and one p orbital hybridize to form two sp hybrid orbitals, with two unhybridized p orbitals used for the triple bond.
- Therefore, the hybridization of carbon in acetylene is **sp**.

### Final Answer:
- The hybridization of carbon in diamond, graphite, and acetylene are respectively **sp³, sp², and sp**.