Using a molecular orbital diagram, predict the bond order, stability and magnetic character of O₂^–,O₂⁺, and  . Also, write their electronic configurations.

O₂^–(Superoxide ion): This ion is formed by the addition of one electron.
O₂ + e^- → O₂
This additional electron will be added up in the  molecular orbital.
Electronic configuration:

Bond order:
Here Nb = 8;   Na = 5

Stability : As the bond order is positive, it is quite stable.
Magnetic character: It has one unpaired electron in the  molecular orbital. 
therefore, it is paramagnetic.
O⁺² ion  This ion is formed by the loss of one electron from O₂ molecule.
O₂ →O₂⁺ e^–
This electron will be lost from *2p_x or *2py molecular orbital.
Electronic configuration:

Bond order:  Here Nb = 8;    Na = 3

Stability: As the bond order is positive, it is quite stable.
Magnetic character: Since  ion has one unpaired electron in the  orbital, therefore, it is paramagnetic.
: The ion is formed when O₂ molecule gains two electrons

These two electrons will be added to the *2p_x and *2py molecular orbitals, one in each.
Electronic configuration:

Bond order: Here N_b = 8;    N_a = 6

Stability: As the bond order is positive, it is quite stable.
Magnetic character: As one of the molecular orbital has a unpaired electron, it is diamagnetic.