Question

Write the electronic configuration of oxygen molecule on the basis of Molecular Orbital Theory. Justify the presence of double bond in it and account for its paramagnetic character​

Answer 1

Answer:

Thus, oxygen molecule has two bonds. i.e., one is bond and one p bond. The last two electrons in p2px∙ and p2py∙ orbitals will remain unpaired. Therefore, oxygen molecule has paramagnetic character due to the presence of two unpaired electrons.

Answer 2

Answer:

Electronic configuration of O₂ molecule:

$\sigma1s^{2}\;\sigma^{*}1s^{2}\;\sigma2s^{2}\; \sigma^{*}2s^{2}\;\sigma2p^{2}\;\pi 2p_{x}^{2}\;\pi 2p_{y}^{2}\;\pi ^{*}2p_{x}^{1}\; \pi _{*}2p_{y}^{1}$

The bond order of   O₂ is two and two unpaired electrons responsible for paramagnetic character.

Explanation:

We know that the electronic configuration of oxygen atom is:

1s²2s²2p⁴

Therefore, the one atom atom has 8 electrons then in a oxygen molecule the two oxygen atoms will have 16 electrons.

On the basis of molecular orbital theory: the electronic configuration of O₂ molecule is:

$\sigma1s^{2}\;\sigma^{*}1s^{2}\;\sigma2s^{2}\; \sigma^{*}2s^{2}\;\sigma2p^{2}\;\pi 2p_{x}^{2}\;\pi 2p_{y}^{2}\;\pi ^{*}2p_{x}^{1}\; \pi _{*}2p_{y}^{1}$

Bond order can be calculated by halved the difference of electrons in antibonding and bonding molecular orbital.

$Bond\; order=\frac{1}{2}[10-6]=\frac{4}{2} =2$

Therefore, the presence of double bond between oxygen atoms in O₂ molecule. The paramagnetic character is due to the presence of two unpaired electrons  $\pi ^{*}2p_{x}$  and  $\pi ^{*}2p_{y}$  in molecular orbitals of O₂ molecule.