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Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2023 / Paper 1 / Section A(iii) / 5 / i / a / Solution 2026-10-20

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The oxygen atom has $1 s^{2} 2 s^{2} 2 p^{4}$ . Each oxygen contributes six valence electrons, so the molecular diagram contains twelve. Taking the internuclear direction as

z

, the relevant ordering for oxygen is

σ_{g} (2 s) < σ_{u}^{*} (2 s) < σ_{g} (2 p_{z}) < π_{u} (2 p_{x}) = π_{u} (2 p_{y}) < π_{g}^{*} (2 p_{x}) = π_{g}^{*} (2 p_{y}) < σ_{u}^{*} (2 p_{z}) .

(1)

The valence occupation is

[σ_{g} (2 s)]^{2} [σ_{u}^{*} (2 s)]^{2} [σ_{g} (2 p_{z})]^{2} [π_{u} (2 p_{x})]^{2} [π_{u} (2 p_{y})]^{2} [π_{g}^{*} (2 p_{x})]^{1} [π_{g}^{*} (2 p_{y})]^{1} .

(2)

The final two electrons occupy the degenerate antibonding orbitals singly with parallel spin, following Hund's rules. Thus oxygen has two unpaired electrons and is paramagnetic. The

1 s

core orbitals are deliberately omitted from the valence-only molecular orbital diagram.

Figure 1.
Valence molecular-orbital occupation of ground-state oxygen
. Original qualitative energy-level diagram with both oxygen atomic-orbital sets and all valence molecular orbitals. Energies and separations are illustrative; arrows indicate electron occupation. The two singly occupied antibonding pi orbitals explain the triplet ground state. **Details**.

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