 Solution

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 $s{p}^2$ to tetrahedral $s{p}^3$, removing its effective C–N pi bonding. The product is described by ${\mathrm{H}}_3\mathrm{C}-{\mathrm{N}}^{+}\equiv \mathrm{N}$: 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.