Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / 2 / c

One consistent localized orbital hybridization description uses the Lewis structuresFor methyldiazonium, take C as $s{p}^3$ and both N atoms as $sp$. For diazomethane, the double-bond resonance contributor suggests C as $s{p}^2$, central N as $sp$, and terminal N as $s{p}^2$. The central N uses two opposite $sp$ hybrids for the linear C–N–N sigma bond framework. In the localized double-bond picture, terminal N uses one $s{p}^2$ hybrid for its sigma bond, two for lone pairs, and its remaining p atomic orbital for a pi bond.

These labels specify an approximate bonding construction, not uniquely measurable atomic properties. An alternative representation takes terminal N as $sp$, with one $sp$ lone pair and one p lone pair; the latter can participate in the delocalized C–N–N $\pi$ system. It describes the same resonance more explicitly. Neither construction requires an unpaired electron or a bent central N.

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) isHere $c,t$ 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.

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.