= Solution
\b[On carbon <protonation>, the C–N <bond length> increases and the N–N <bond length> decreases.] Forming the fourth C <sigma bond> changes the carbon from approximately planar $sp^2$ to tetrahedral $sp^3$, removing its effective C–N <pi bonding>. The product is described by $\mathrm{H_3C-N^+\equiv N}$: C–N becomes predominantly single and N–N predominantly triple. The <resonance> of the unprotonated species makes its actual <bond orders> intermediate rather than exact integers.
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