Question

According to molecular orbital theory which of the following is correct?

• A. LUMO level for $C_2$ molecule is ${\sigma }_{2p_z}$orbital
• B. In $C_2$ molecules both the bonds are $\pi$ bonds
• C. In $C_2^{2-}$ ion there is one $\sigma$ and two $\pi$ bonds
• D. All the above are correct

Answer 1

The correct option is D All the above are correct

Configuration for $C_2=\sigma 1s^2<{\sigma }^{\ast }1s^2<\sigma 2s^2<{\sigma }^{\ast }2s^2<\underset{HOMO}{\underset{}{\pi 2p_y^2=\pi 2p_x^2}}<\underset{LUMO}{\sigma 2p_z}$
There is no sigma bond in the $C_2$ molecule because the sigma bonding electrons are cancelled by the sigma antibonding electrons and only pi bonding electrons left.
It is important to note that double bond in $C_2$ consists of both pi bonds because of the presence of four electrons in two pi molecular orbitals.

LUMO of the $C_2$ is ${\sigma }_{2p_z}$orbital.

Configuration for $C_2^{2-}=\sigma 1s^2<{\sigma }^{\ast }1s^2<\sigma 2s^2<{\sigma }^{\ast }2s^2<\underset{HOMO}{\underset{}{\pi 2p_y^2=\pi 2p_x^2}}<\underset{LUMO}{\sigma 2p_z^2}$
In $C_2^{2-}$, $\sigma 2p_z$ also gets filled. Therefore, in $C_2^{2-}$ ion there is one $\sigma$ and two $\pi$ bonds.