Question

Calculate the sum of ratios of $\sigma$ bonding electrons and $\pi$ bonding electrons in $C_2$ molecule.

Answer 1

The electronic configuration of $C_2$ molecule is as given below.
$C_2={\sigma }_{\left(1s\right)}^2{\sigma }_{\left(1s\right)}^{\ast 2}{\sigma }_{\left(2s\right)}^2{\sigma }^{\ast 2}\left(2s\right)\pi 2p_y^2=\pi 2p_z^2$
$C_2$ molecule contains 4 sigma bonding electrons, 4 sigma antibonding electrons and 4 pi electrons.
$Ratio=\frac{Numberof\sigma bondingelectrons}{Numberof\pi bondingelectrons}=\frac{4}{4}=1$
Hence, the ratio of $\sigma$ bonding electrons and $\pi$ bonding electrons in $C_2$ molecule is 1:1.