Question

Explain the formation of $O_2$ molecule using molecular orbital theory.

Answer 1

Oxygen molecule $O_2$: The electronic configuration of oxygen $(z=8)$ in the ground state is $1s^{2}2s^{2}2p^4$. Each oxygen atom has $8$ electrons, hence in $O_2$ molecule there are $16$ electrons.

Therefore, the electronic configuration of $O_2$ is as follows :

$O_2:KK{\left({\sigma }_{2s}\right)}^2{\left({{\sigma }_{2s}}^{\bullet }\right)}^2{\left({{\pi }_{2p}}_x\right)}^2={\left({{\pi }_{2p}}_y\right)}^2{\left({{\pi }_{2p}}_x\right)}^1={\left({{{\pi }_{2p}}_y}^{\bullet }\right)}^1$

The molecular orbital energy level diagram of oxygen molecule is given as follows :

Bond order $\frac{N_b-N_a}{2}=\frac{8-4}{2}=2$
Thus, oxygen molecule has two bonds. i.e., one is bond and one p bond. The last two electrons in $p_{2px}^{\bullet }$ and $p_{2py}^{\bullet }$ orbitals will remain unpaired. Therefore, oxygen molecule has paramagnetic character due to the presence of two unpaired electrons.