Question

Calculate the ratio of $\sigma$ bonding electrons and $\pi$ bonding electrons in $B_2$ molecule.

Answer 1

The electronic configuration of $B_2$ molecule is as given below.

$B_2=\sigma 1s^2{\sigma }^{\ast }\left(1s^2\right)\sigma \left(2s^2\right){\sigma }^{\ast }\left(2s^2\right)\pi 2p_y^1=\pi 2p_z^1$

$B_2$ molecule contains 4 sigma bonding electrons, 4 sigma antibonding electrons and 2 pi electrons.

$Ratio=\frac{Numberof\sigma bonding{e}^{-}}{Numberof\pi bonding{e}^{-}}=\frac{4}{2}=2$

Hence, the ratio of $\sigma$ bonding electrons and $\pi$ bonding electrons in $B_2$ molecule is $2:1$.