 Solution

For oxygen, eight valence electrons occupy bonding orbitals and four occupy antibonding orbitals. Its bond order isThe additional electron in superoxide enters a ${\pi }_g^{\ast }(2p)$ antibonding orbital, givingReduced 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.