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

For a hydrogen-like ion, the Coulomb potential is stronger by the nuclear atomic number $Z$, and the energy levels scale as $Z^2$:Here $Z$ is the number of protons in the atomic nucleus; the ion has exactly one electron, so its net electric charge is $(Z-1)e$. The zero of energy is a separated stationary electron and atomic nucleus at infinite distance. This is the nonrelativistic hydrogenic approximation using the supplied $R_H$; a more precise isotope-specific result also corrects the reduced mass.

The lowest two energy levels have $n=1$ and $n=2$. For carbon, $Z=6$, even though the ion's net charge is $5+$:Increasing $n$ raises the energy toward zero; the most negative value is the ground-state energy.

An emitted photon has positive energyFor spontaneous emission into $n_f=2$, the conservation of energy equation isConsequentlyThe principal quantum number must be an integer, so the transition is $\boxed{n_i=3⟶n_f=2}$. Its calculated wavelength is approximately $18.22\mathrm{nm}$, agreeing with the rounded observation.

End-on overlap produces $\sigma$ and ${\sigma }^{\ast }$ molecular orbitals; side-on overlap produces $\pi$ and ${\pi }^{\ast }$ molecular orbitals. Matching signs of the facing atomic orbital lobes give constructive interference and a bonding molecular orbital. Opposite signs give destructive interference and an additional nodal plane between the atomic nuclei in the antibonding molecular orbital.

Take the internuclear direction as $x$, and define both atomic $p_x$ functions as positive on their own right-hand side. Up to wavefunction normalization, the combinations areThe minus sign in the bonding $p_x$ combination follows from this common-axis quantum phase convention: the facing lobes of the two unmodified $p_x$ functions have opposite signs. It does not make that combination antibonding. Every $\pi$ molecular orbital also has a nodal plane containing the internuclear axis; the additional midpoint node distinguishes ${\pi }^{\ast }$ from $\pi$.

The labels apply when the molecule has an inversion centre, as a homonuclear diatomic molecule does. The label $g$ means $\psi (-\mathbf{r})=\psi (\mathbf{r})$, and $u$ means $\psi (-\mathbf{r})=-\psi (\mathbf{r})$, with the origin at that centre. In a heteronuclear diatomic molecule, inversion exchanges different atomic nuclei and these are generally not valid symmetry labels.

For the p-derived combinations above,Inversion both exchanges the two centres and changes the sign of each individual p atomic orbital. Applying both operations to each sum or difference gives these labels. The star denotes antibonding molecular orbitals, independently of parity: a bonding molecular orbital need not always be $g$.

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.