= Solution
\b[There are $\boxed{4+2(1)+2(5)=16}$ <valence electrons>, all paired, so <diazomethane> exhibits <diamagnetism>.] A localized orbital occupancy consistent with the double-bond construction in 2(c)(i) is
$$
\begin{array}{l|r|l}
\text{occupied orbitals}&\text{electrons}&\text{atomic-orbital ingredients}\\\hline
2\ \mathrm{C-H}\ \sigma&4&\mathrm C(sp^2)+\mathrm H(1s)\\
\mathrm{C-N}\ \sigma&2&\mathrm C(sp^2)+\mathrm N_c(sp)\\
\mathrm{N-N}\ \sigma&2&\mathrm N_c(sp)+\mathrm N_t(sp^2)\\
\mathrm{C-N}\ \pi&2&\mathrm C(2p)+\mathrm N_c(2p)\\
\mathrm{N-N}\ \pi&2&\mathrm N_c(2p_\perp)+\mathrm N_t(2p)\\
2\ \mathrm N_t\text{ lone pairs}&4&2\ \mathrm N_t(sp^2)
\end{array}
$$
Here $c,t$ mean central and terminal. Each listed occupied spatial orbital contains two <electrons> of opposite <spin>. The central N's two p <atomic orbitals> are perpendicular, so they can support the two perpendicular <pi bonds>. No central-N <lone pair> is left over: its positive <formal charge> accompanies four bonding pairs in this <Lewis structure>.
Allowing <resonance> mixes a terminal-N <lone pair> into the C–N–N $\pi$ orbitals. In the alternative terminal-$sp$ basis, the same electron count is four occupied $\sigma$ orbitals ($8$ electrons), one terminal-$sp$ <lone pair> ($2$), one perpendicular N–N $\pi$ orbital ($2$), and two occupied orbitals of a three-centre C–N–N $\pi$ system ($4$). The $16$-electron count and the paired ground state survive this change of description; the localized picture does not assert two perfectly isolated double bonds.
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