Question

Nitric oxide, NO, is an odd-electron species. In the solid-state, it dimerizes to form N₂O₂. The most stable isomer is the symmetric dimer ONNO.

a) Draw three resonance structures for the dimer.
b) Determine the formal charges on each atom in each resonance structure.
c) Circle the resonance structure that best represents the bonding in ONNO.
d) If ONNO is reduced by two electrons to make N₂O₂²⁻, what happens to the N-N and N-O bond lengths?
e) Draw the N₂O₂²⁻ ion, indicating a dipole moment of zero.

Answer 1

The reduction of ONNO to form N₂O₂²⁻ would likely increase the N-N bond length and decrease the N-O bond lengths, reflecting the changes in bonding interactions caused by the added electrons. The N₂O₂²⁻ ion would have equal N-O bond lengths and a symmetric structure, resulting in a zero dipole moment.

Step-by-step explanation:

When the symmetric dimer ONNO is reduced by two electrons to form N₂O₂²⁻, the N-N bond length is expected to increase, while the N-O bond lengths would likely decrease. This prediction is based on the electrons' reduction causing a change in the bonding. Upon reduction, the additional electrons would likely occupy antibonding orbitals, thereby weakening the N-N bond, leading to an increase in bond length. In contrast, additional electrons in N-O π* antibonding orbitals would decrease the bond order, making these formerly double bonds resemble more like 1.5 bonds, hence shortening these bonds compared to the oxidized form.

Drawing the N₂O₂²⁻ ion with a dipole moment of zero involves showing two NO⁻ units joined by two shared electrons. Each of the NO⁻ units would have a bond order of 1.5 to reflect equal N-O bond lengths, resonating between single and double bonded structures. Such a structure would be symmetric, canceling out any possible dipole moments across the molecule.