= Solution
The framework has a planar $\mathrm{H_2C}$ group attached to a straight C–N–N axis. Two useful octet <resonance contributors> are
$$
\boxed{\mathrm{H_2C=N^+=N^-\ \leftrightarrow\ H_2C^--N^+\equiv N}.}
$$
In the left contributor, the terminal <nitrogen> has two <lone pairs>; in the right, the terminal <nitrogen> and <carbon> each have one. Each formula has 16 valence <electrons> and zero net <electric charge>. The double-headed resonance arrow changes the electronic allocation while retaining the same nuclear framework; these are not two different equilibrium species.
Back to article page
