The double bond in $C_{2}$ molecule consist of:

Question

Accepted Answer

The double bond in the $C_{2}$ molecule consists of both $\pi$ bonds. Let's explore why Option 2 is correct and why the other options are incorrect.

### Explanation for Option 2 (Correct Answer):
1. **Electronic Configuration of Carbon**: Carbon has an electronic configuration of $1s^2 2s^2 2p^2$. In the $C_{2}$ molecule, each carbon atom contributes two electrons from the 2p orbitals for bonding.
   
2. **Bond Formation in $C_{2}$**: In the $C_{2}$ molecule, the two carbon atoms form a double bond. However, unlike typical double bonds that consist of one $\sigma$ bond and one $\pi$ bond, the $C_{2}$ molecule is unique. The two carbon atoms in $C_{2}$ form two $\pi$ bonds using their 2p orbitals. This is due to the fact that the 2s orbitals do not participate in $\sigma$ bonding in this particular molecule, leading to the formation of two $\pi$ bonds instead.

3. **Molecular Orbital Theory**: According to molecular orbital theory, the $C_{2}$ molecule does not have a $\sigma$ bond in its ground state. The bonding involves two $\pi$ bonds formed by the overlap of p orbitals, which is consistent with the observed properties of the $C_{2}$ molecule.

### Explanation for Why Other Options are Incorrect:

1. **Option 1 (both $\sigma$ bonds)**:
   - This option is incorrect because $\sigma$ bonds are typically formed by the head-on overlap of orbitals, such as s-s, s-p, or p-p orbitals. In the case of $C_{2}$, the bonding involves side-by-side overlap of p orbitals, resulting in $\pi$ bonds, not $\sigma$ bonds.

2. **Option 3 (one $\sigma$ and one $\pi$ bond)**:
   - This is the typical configuration for double bonds in many molecules, but not for $C_{2}$. In $C_{2}$, the absence of a $\sigma$ bond is due to the specific electronic configuration and the way the p orbitals overlap, leading to the formation of two $\pi$ bonds instead.

3. **Option 4 (None of the above)**:
   - This option is incorrect because it contradicts the established understanding of the bonding in $C_{2}$. The correct and supported explanation is that the double bond in $C_{2}$ consists of two $\pi$ bonds, as described in Option 2.

In conclusion, the unique bonding situation in the $C_{2}$ molecule, where the double bond consists of two $\pi$ bonds, is accurately described by Option 2. This explanation aligns with molecular orbital theory and the known properties of the $C_{2}$ molecule.