 Solution

An internally consistent elementary orbital hybridisation model is $s{p}^2$ at carbon and $sp$ at both nitrogen atoms; the hydrogens use $1s$ orbitals. The three carbon hybrids form two C–H sigma bonds and one C–N sigma bond, leaving one perpendicular p orbital. The central nitrogen uses opposite $sp$ hybrids for its two collinear sigma connections, leaving two perpendicular p orbitals for the two pi systems.

At terminal nitrogen, one $sp$ hybrid points toward the central nitrogen and the other houses an axial lone pair. One perpendicular p orbital forms the second, N–N-only pi system; the other participates in the three-atom C–N–N pi system. This description remains the same across both resonance contributors; counting lone-pair domains in just one drawing and assigning a different terminal hybridisation to each contributor would obscure the delocalization. orbital hybridisation is a useful choice of localized basis, not a uniquely measurable label: advanced treatments of diazomethane require a superposition of bonding patterns.^[ref]