= Solution
An internally consistent elementary <orbital hybridisation> model is \b[$sp^2$ 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.<On the electronic structure of the diazomethane molecule by André G. H. Barbosa and João G. S. Monteiro>[Abstract: limitations of single bonding patterns and multiconfigurational description]{ref}
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