Write the molecular orbit configuration of o2+. Calculate its bond order and predict its magnetic bahaviour
Answers
In a O_{2}^{+} molecule, there are total 15 electrons. The molecular orbital configuration O_{2}^{+} is as follows.
\sigma 1s^{2} \sigma^{*} 1s^{2} \sigma 2s^{2}\sigma^{*} 2s^{2} \sigma 2p_{x}^{2} \pi 2p_{x}^{2} \pi 2p_{y}^{2} \pi^{*} 2p_{x}^{1} \pi^{*} 2p_{y}^{0}
The formula for bond order is as follows.
Bond order = \frac{1}{2} [Bonding - Non bonding]
There are 10 bonding and 5 non-bonding electrons in the orbitals according to the molecular orbital configuration.
Therefore, Bond order = 1/2 [Bonding - Non bonding]
= 1/2 [10 - 5]
= 1/2 [5]
= 2.5
Thus, the bond order of O_{2}^{+} is 2.5.