Carbon atoms in `C_2(CN)_4` are :

Options

A. `sp`-hybridsed
B. `sp^2 -`hybrdised
C. `sp` and` sp^2 `-hybridesed
D. `sp, sp^2` and `sp^3` -hubridesed

Explanation

To determine the hybridization states of the carbon atoms in the compound \( C_2(CN)_4 \), we will analyze the structure step by step.

### Step-by-Step Solution:

1. **Identify the Structure**:
- The formula \( C_2(CN)_4 \) indicates that there are 2 carbon atoms and 4 cyanide (CN) groups.
- Each cyanide group consists of a carbon atom triple-bonded to a nitrogen atom (C≡N).

2. **Draw the Structure**:
- The structure can be represented as:
\[ C \equiv N - C \equiv N - C \equiv N - C \equiv N \]
- Here, the two carbon atoms are connected to four cyanide groups.

3. **Count the Carbon Atoms**:
- In total, we have 2 carbon atoms from the \( C_2 \) part and 4 carbon atoms from the 4 cyanide groups, giving us a total of 6 carbon atoms.

4. **Determine the Hybridization**:
- For each carbon atom in the cyanide groups (C≡N), the carbon is involved in a triple bond with nitrogen. A triple bond consists of one sigma bond and two pi bonds.
- The carbon atom in a triple bond uses one \( s \) orbital and two \( p \) orbitals for hybridization, resulting in \( sp \) hybridization.

5. **Analyze the Central Carbon Atoms**:
- The two central carbon atoms that are connected to the cyanide groups also form triple bonds with the nitrogen atoms.
- Therefore, these carbon atoms are also \( sp \) hybridized.

6. **Conclusion**:
- All carbon atoms in \( C_2(CN)_4 \) are \( sp \) hybridized.

### Final Answer:
All carbon atoms in \( C_2(CN)_4 \) are \( sp \) hybridized.