What is the hybrid state of carbon in ethyne, graphite and diamond ?

Options

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

Explanation

To determine the hybrid state of carbon in ethyne, graphite, and diamond, we will analyze each form of carbon step by step.

### Step 1: Hybridization in Ethyne (C₂H₂)
1. **Structure of Ethyne**: Ethyne is an alkyne with the formula C₂H₂.
2. **Bonding**: In ethyne, each carbon atom forms one sigma bond and two pi bonds with the other carbon atom.
3. **Electron Configuration**: The ground state electron configuration of carbon is 1s² 2s² 2p². In the excited state, one electron from the 2s orbital is promoted to the 2p orbital, resulting in the configuration: 1s² 2s¹ 2p³.
4. **Hybridization**: For the formation of one sigma bond, one s orbital and one p orbital hybridize to form two sp hybrid orbitals. The remaining two p orbitals form the two pi bonds.
5. **Conclusion**: The hybridization of carbon in ethyne is **sp**.

### Step 2: Hybridization in Graphite
1. **Structure of Graphite**: Graphite is a planar structure where each carbon atom is bonded to three other carbon atoms.
2. **Bonding**: Each carbon atom forms three sigma bonds and has one delocalized pi bond due to the free p orbital.
3. **Electron Configuration**: Similar to ethyne, the carbon atoms in graphite undergo hybridization. The three sigma bonds involve three sp² hybrid orbitals, while the remaining p orbital is used for delocalized bonding.
4. **Hybridization**: The hybridization of carbon in graphite is **sp²**.

### Step 3: Hybridization in Diamond
1. **Structure of Diamond**: Diamond has a tetrahedral structure where each carbon atom is bonded to four other carbon atoms.
2. **Bonding**: Each carbon atom forms four sigma bonds, utilizing all four valence electrons.
3. **Electron Configuration**: In diamond, all four orbitals (one s and three p) hybridize to form four sp³ hybrid orbitals.
4. **Hybridization**: The hybridization of carbon in diamond is **sp³**.

### Final Summary
- **Ethyne**: sp hybridization
- **Graphite**: sp² hybridization
- **Diamond**: sp³ hybridization

### Answer
The hybrid states of carbon in ethyne, graphite, and diamond are:
- Ethyne: sp
- Graphite: sp²
- Diamond: sp³

The correct option is **B: sp, sp², sp³**.

---