Question

The two compounds nitrogen dioxide and dinitrogen tetraoxide are introduced in Section 3.13.

(a)${\mathbf{NO}}_{\mathbf{2}}$ is an odd-electron compound. Draw the best Lewis diagrams possible for it, recognizing that one atom cannot achieve an octet configuration. Use formal charges to decide whether that should be the (central) nitrogen atom or one of the oxygen atoms.

(b) Draw resonance forms for${\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}$ that obey the octet rule. The two N atoms are bonded in this molecule.

Short answer

a).

b).

Step-by-step solution

(a) The best Lewis diagrams possible for nitrogen oxide.

The resonance form with all the atoms following octet is;

The resonance hybrid is;

Now, if one atom cannot achieve an octet configuration, then the resonating structures go as;

The unpaired electron that is involved in forming a bond is on the nitrogen atom.

Here, a formal charge of +1 is present on the N-atom, and a formal charge of -1 on one of the terminal oxygen atoms.

The central N-atom has 7 electrons and therefore has not achieved an octet configuration.

(b) The resonance forms for dinitrogen tetraoxide that obey the octet rule.

The ${\mathrm{NO}}_2$dimerizes to form molecule ${\mathrm{N}}_2{\mathrm{O}}_4$, with the two N atoms bonded in all the resonating structures.

The different resonance forms for ${\mathrm{N}}_2{\mathrm{O}}_4$that obey the octet rule is shown below.