Question

Write the molecular orbital configuration of o2 and o2+ calculate the bond order and predict their magnetic behaviour

Answer 1

O2+ is more stable than O2-.
Reason: According to molecular orbital theory O2+has 15 electrons &it has one electron in antibonding orbital.
molecular orbital diagram of  O2+                                                
                                                                                                           

Electronic configuration of O2+


In the case of O2- 17 electrons are present &3 electrons are present in antibonding orbitals. If number of electrons more in antibonding orbital the molecule become unstable.
molecular orbital diagram of O2-


Electronic configuration of O2-

Answer 2

Explanation:

In a $O_{2}$ molecule, there are total 16 electrons. The molecular orbital configuration of $O_{2}$ molecule is as follows.

    $\sigma 1s^{2} \sigma^{*} 1s^{2} \sigma 2s^{2}\sigma^{*} 2s^{2} \sigma 2p_{x}^{2} \pi 2p_{x}^{2} \pi 2p_{y}^{2} \pi^{*} 2p_{x}^{1} \pi^{*} 2p_{y}^{1}$

The formula for bond order is as follows.

            Bond order = $\frac{1}{2} [Bonding - Non bonding]$

There are 10 bonding and 6 non-bonding electrons in the orbitals according to the molecular orbital configuration.

Therefore,     Bond order = $\frac{1}{2} [Bonding - Non bonding]$                      

                                         = $\frac{1}{2} [10 - 6]$

                                         = $\frac{1}{2} [4]$

                                         = 2

Thus, the bond order of $O_{2}$ is 2.

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 = $\frac{1}{2} [Bonding - Non bonding]$                      

                                         = $\frac{1}{2} [10 - 5]$

                                         = $\frac{1}{2} [5]$

                                         = 2.5

Thus, the bond order of $O_{2}^{+}$ is 2.5.