 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-2a/section-b/23/a/solution

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Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 2A / Section B / 23 / a / Solution by

Codex  0 2026-10-20

Choose x along O–O. The valence orbital order appropriate to O

_{2}

is approximately

2 σ_{g} (2 s) < 2 σ_{u}^{*} (2 s) < 3 σ_{g} (2 p_{x}) < 1 π_{u} (2 p_{y}, 2 p_{z}) < 1 π_{g}^{*} (2 p_{y}, 2 p_{z}) < 3 σ_{u}^{*} (2 p_{x}) .

(1)

Its twelve valence electrons occupy

(2 σ_{g})^{2} (2 σ_{u}^{*})^{2} (3 σ_{g})^{2} (1 π_{u})^{4} (1 π_{g, y}^{*})^{1} (1 π_{g, z}^{*})^{1} .

(2)

The two degenerate pi-antibonding orbitals are singly occupied with parallel spins according to Hund's rules. Thus $S = 1$ , multiplicity $2 S + 1 = 3$ , with three possible

M_{S}

values. A triplet is a many-electron spin classification, not three electrons in one orbital. The resulting ground term is

^{3} Σ_{g}^{-}

. Exchange stabilization favours this triplet over the singlets formed within the same frontier-orbital manifold.

Figure 1.
Original valence-O2 diagram with the two singly occupied pi-antibonding orbitals
. **Details**.

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