= Solution
For oxygen, eight valence electrons occupy bonding orbitals and four occupy antibonding orbitals. Its <bond order> is
$$
B(\mathrm O_2)=\tfrac12(8-4)=2.
$$
The additional electron in <superoxide> enters a $\pi_g^*(2p)$ <antibonding orbital>, giving
$$
\boxed{B(\mathrm O_2^-)=\tfrac12(8-5)=1.5}.
$$
Reduced bond order weakens and lengthens the bond, consistent with the longer bond in superoxide. The diagram explains the direction of the change; it does not predict the two numerical bond lengths without a quantitative electronic-structure calculation.
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