Using MO theory predict which of the following species has the shortest bond length?

According to Molecular Orbital Theory, bond order is directly proportional to bond strength and inversely proportional to bond length:

$$\text{Bond Length} \propto \frac{1}{\text{Bond Order}}$$

A higher bond order indicates more shared electron density holding the nuclei tightly together, resulting in a shorter bond distance.

The base neutral molecule $$\text{O}_2$$ contains $$16$$ electrons. Its electronic configuration following the valence shell energy ordering is:

$$\sigma_{1s}^2 \ \sigma_{1s}^{*2} \ \sigma_{2s}^2 \ \sigma_{2s}^{*2} \ \sigma_{2p_z}^2 \ \left(\pi_{2p_x}^2 = \pi_{2p_y}^2\right) \ \left(\pi_{2p_x}^{*1} = \pi_{2p_y}^{*1}\right)$$

The formula to calculate the bond order is:

$$\text{Bond Order} = \frac{N_b - N_a}{2}$$

Where $$N_b$$ is the number of bonding electrons and $$N_a$$ is the number of antibonding electrons. Let's analyze each ion:

• Option A: $$\text{O}_2^{2+}$$ ($$14$$ electrons)

  Two electrons are removed from the highest occupied anti-bonding molecular orbitals ($$\pi^*_{2p}$$).

  $$\text{Bond Order} = \frac{10 - 4}{2} = 3$$



• Option B: $$\text{O}_2^+$$ ($$15$$ electrons)

  One electron is removed from an anti-bonding orbital ($$\pi^*_{2p}$$).

  $$\text{Bond Order} = \frac{10 - 5}{2} = 2.5$$



• Option C: $$\text{O}_2^-$$ ($$17$$ electrons)

  One extra electron is added to an anti-bonding orbital ($$\pi^*_{2p}$$).

  $$\text{Bond Order} = \frac{10 - 7}{2} = 1.5$$



• Option D: $$\text{O}_2^{2-}$$ ($$18$$ electrons)

  Two extra electrons are added to anti-bonding orbitals ($$\pi^*_{2p}$$).

  $$\text{Bond Order} = \frac{10 - 8}{2} = 1$$

Comparing the calculated values, $$\text{O}_2^{2+}$$ has the highest bond order ($$3$$), which means it possesses the tightest chemical bond and consequently the shortest bond length.

Answer: Option A — $$\text{O}_2^{2+}$$