Question

Calculate the ratio of $\sigma $ bonding electrons and $\pi $ bonding electrons in ${{B}_{2}}$ molecule.

Answer 1

Hint: Think about the molecular orbital theory. How electrons go into the bonding and antibonding orbitals of the molecules and in what order.

Complete answer:
Here we have to calculate the ratio of the $\sigma $ bonding electrons and $\pi $ bonding electrons. Thus, we only need to take into consideration the electrons that are present in the bonding orbitals of the molecule and not the ones in the antibonding orbitals.
To calculate this, we need to first understand how many electrons are present in the ${{B}_{2}}$ molecule. This molecule has no charge so the number of electrons present will be the sum of the electrons present in 2 atoms of boron.
The atomic number of boron is 5. Hence the number of electrons present in the ${{B}_{2}}$ molecule is 10.
Now, we will write the electronic configuration of the molecule.
${{B}_{2}}=\sigma 1{{s}^{2}}\cdot \sigma *1{{s}^{2}}\cdot \sigma 2{{s}^{2}}\cdot \sigma *2{{s}^{2}}\cdot \pi 2{{p}_{x}}^{1}=\pi 2{{p}_{y}}^{1}\cdot \sigma 2{{p}_{z}}^{0}$
The anti-bonding orbitals are marked with a $*$. All the 2p antibonding orbitals are empty. So now we can calculate the ratio.
No. of electrons in $\sigma $bonding orbitals = 4
These electrons are present in the $1s$ and $2s$ bonding orbitals.
No. of electrons in $\pi $ bonding orbitals = 2
These electrons are present in the $2{{p}_{x}}$ and $2{{p}_{y}}$ bonding orbitals.
So, we can calculate the ratio to be:
\[Ratio=\dfrac{4}{2}\]

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

Note: Please do not get confused between the electrons that are present in the bonding orbitals and the electrons that help in the bonding of the 2 atoms. The electrons in the bonding orbitals are all the electrons that are present between the 2 nuclei of the atoms involved.