Superoxide ( ) and nitric oxide ( ) are both odd-electron species
CHEMISTRY · Single correct · CBSE · Chemistry · Chemical Bonding and Molecular Structure · Chemical Bonding and Molecular Structure
Problem
Superoxide (O2−) and nitric oxide (NO) are both odd-electron species. Which of the following best accounts for the observation that the O−O bond length in O2− is intermediate between those in O2 and O22−, whereas the electron density in NO is unevenly distributed between the two atoms?
(a) In O2−, each oxygen can employ empty d-orbitals to accommodate the extra electron, giving an averaged bond length; in NO the lack of accessible d-orbitals localises the unpaired electron and creates asymmetry.
(b) In O2−, the unpaired electron is delocalized equally over two identical oxygen atoms, giving a bond order of 1.5 and an intermediate O−O bond length; in NO the heteronuclear nature causes the unpaired electron to reside mainly on the less electronegative nitrogen, resulting in a bond order of 2.5 and uneven electron density.
(c) Adding an electron to O2 raises its bond order to 2, so the O−O bond in O2− stays as short as in O2; in NO the odd electron lowers the bond order to 1.5, weakening and deforming the bond.
(d) Because nitrogen and oxygen have nearly equal electronegativities, resonance in NO spreads the unpaired electron evenly over both atoms, whereas in O2− the negative charge pulls electron density toward a single oxygen atom, producing an intermediate bond length.