Q.Using molecular orbital theory, compare the bond energy and magnetic character of O2^+ and O2^- species.

Concept: Bond Order and Magnetic Properties

Molecular orbital theory predicts that bond energy increases with bond order, while magnetic character depends on unpaired electrons.

For O₂ (16 electrons), the MO configuration is:

For O₂⁺ (15 electrons): Remove one electron from the antibonding π* orbital.

• Bond order = $\frac{10 - 5}{2} = 2.5$
• Unpaired electrons = 1 → paramagnetic

For O₂⁻ (17 electrons): Add one electron to the antibonding π* orbital. …

Why Molecular Orbital Theory Reveals Bond Strength and Magnetism

Molecular orbital theory distributes electrons across bonding and antibonding orbitals. The bond order—half the difference between bonding and antibonding electrons—directly predicts bond strength: higher bond order means a stronger, shorter bond. Magnetism depends on unpaired electrons: any species with unpaired electrons is paramagnetic, while fully paired electrons yield diamagnetism.

For oxygen and its ions, the key is tracking how adding or removing electrons changes the occupancy of the frontier orbitals.

Step-by-Step Comparison

1. Write the molecular orbital configuration for neutral $\mathrm{O}_2$

Oxygen has 8 electrons per atom, so $\mathrm{O}_2$ has 16 electrons total. The MO filling order for $\mathrm{O}_2$ (a second-row homonuclear diatomic with $Z \leq 7$ showing $\sigma$-$\pi$ mixing, though for $\mathrm{O}_2$ the order is actually standard) is:

The last four electrons occupy the $\pi_{2p}$ bonding pair (4 electrons) and the $\pi_{2p}^*$ antibonding pair (2 electrons, one in each degenerate orbital by Hund's rule).

2. Calculate bond order and magnetic character of $\mathrm{O}_2$

Bond order for $\mathrm{O}_2$:

The two unpaired electrons in $\pi_{2p_x}^*$ and $\pi_{2p_y}^*$ make $\mathrm{O}_2$ paramagnetic.

3. Determine the configuration of $\mathrm{O}_2^+$ (15 electrons)

Removing one electron from $\mathrm{O}_2$ takes it from an antibonding $\pi^*$ orbital:

Bond order:

One unpaired electron remains, so $\mathrm{O}_2^+$ is paramagnetic.

4. Determine the configuration of $\mathrm{O}_2^-$ (17 electrons)

Adding one electron to $\mathrm{O}_2$ places it in an antibonding $\pi^*$ orbital, pairing one of the previously unpaired electrons: …