= Solution
The four <sigma bonds> are the two C–H connections, C–N and N–N, containing eight electrons. A terminal-nitrogen axial <lone pair> contains two more. Of the remaining six electrons, two occupy an N–N <pi bond> lying in the molecular plane, while four occupy a perpendicular <three-centre four-electron pi bond> extending over C–N–N. There are therefore 16 <valence electrons> in total.
For orientation, a three-site model with equal on-site energy $\alpha$ and adjacent interaction $\beta<0$ gives perpendicular pi levels at $\alpha+\sqrt2\beta$, $\alpha$, and $\alpha-\sqrt2\beta$. The lower two are doubly occupied and the upper is empty. The middle level has a terminal-end combination with little central-atom amplitude; unequal C and N orbital energies change its coefficients and make it only approximately nonbonding.
The \b[<highest occupied molecular orbital> is the upper occupied perpendicular pi orbital], with substantial <carbon> amplitude, broadly above the strongly bonding sigma framework and the lower occupied pi orbitals. It is not the terminal axial <nitrogen> <lone pair>. Real <diazomethane> has C–N bonding and N–N antibonding character in this frontier orbital, rather than being an exactly nonbonding equal-site model. Photoelectron and Rydberg-state experiments support this assignment.<Vibronic structure and ion core interactions in Rydberg states of diazomethane: an experimental and theoretical investigation by Igor Fedorov and others>[Abstract and discussion of ionization: removal from a C–N-bonding, N–N-antibonding pi orbital]{ref}
\Image[/past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1-diazomethane-orbitals.png]
{title=Original qualitative decomposition of diazomethane into four sigma bonds, an axial lone pair and two perpendicular pi systems}
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