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

Question

Accepted Answer

The question asks about the nature of the double bond in $C_{2}$, and the correct answer is given as Option 2: Both $\pi$ bonds. Let's provide a detailed explanation supporting this answer and explain why the other options are incorrect.

### Explanation Supporting Option 2:

In the molecule $C_{2}$, the double bond is indeed composed of two $\pi$ bonds. This is a unique and somewhat unconventional bonding situation. In typical double bonds, such as those found in ethylene ($C_2H_4$), the double bond consists of one $\sigma$ bond and one $\pi$ bond. However, in the case of the diatomic carbon molecule ($C_{2}$), the bonding is different due to the electronic configuration and the nature of carbon atoms.

1. **Electronic Configuration**: Carbon has the electronic configuration of $1s^2 2s^2 2p^2$. In $C_{2}$, each carbon atom contributes two unpaired electrons in the 2p orbitals. These p orbitals can overlap to form $\pi$ bonds.

2. **Bonding in $C_{2}$**: The $C_{2}$ molecule is known to have a bond order of 2, which is achieved through the formation of two $\pi$ bonds. This is supported by molecular orbital theory, which suggests that the $\sigma$ bonding orbital is not occupied in the ground state of $C_{2}$, leading to the formation of two $\pi$ bonds instead.

### Explanation for Why Other Options are Incorrect:

1. **Option 1: Both $\sigma$ bonds** - This option is incorrect because $\sigma$ bonds are formed by the head-on overlap of orbitals, which is not the case in $C_{2}$. The molecule does not have two $\sigma$ bonds; instead, it has two $\pi$ bonds due to the side-by-side overlap of p orbitals.

2. **Option 3: One $\sigma$ and one $\pi$ bond** - This is the typical composition of a double bond in many organic molecules, but not in $C_{2}$. As explained, $C_{2}$ uniquely forms two $\pi$ bonds without a $\sigma$ bond in its ground state.

3. **Option 4: Both coordinate bonds** - Coordinate bonds (or dative covalent bonds) involve one atom providing both electrons for the bond. This is not applicable to $C_{2}$, as the bonding involves mutual sharing of electrons between the two carbon atoms, specifically forming $\pi$ bonds.

In conclusion, the correct answer is Option 2: Both $\pi$ bonds, due to the unique electronic structure and bonding nature of the $C_{2}$ molecule.