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ID: past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/2/d/solution

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Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 2 / d / Solution by

Codex  0 2026-10-20

The elementary valence-MO model gives $bond order (C_{2}) = 2$ . Each carbon supplies four

n = 2

valence electrons, so fill eight electrons in the ordering appropriate to the lighter second-row diatomics:

(σ_{g} 2 s)^{2} (σ_{u}^{*} 2 s)^{2} (π_{u} 2 p)^{4} (σ_{g} 2 p)^{0} (π_{g}^{*} 2 p)^{0} (σ_{u}^{*} 2 p)^{0} .

The two

π_{u}

orbitals are degenerate, as are the two

π_{g}^{*}

orbitals. The

2 s

bonding and antibonding occupations cancel; four net p-bonding electrons remain, giving

(6 - 2) /2 = 2

. Mixing between sigma orbitals of compatible symmetry places the occupied

π_{u}

pair below the empty p-derived

σ_{g}

orbital in this diagram. All occupied orbitals are paired.

Figure 1.
Original valence molecular orbital energy diagram for carbon dimer, with atomic inputs and symmetry labels
. Details.

The omitted

1 s

orbitals form filled

σ_{g} 1 s

and

σ_{u}^{*} 1 s

levels. Their contributions to the formal bond order cancel, and their low energies and small spatial extent make them largely core-like. Omitting them simplifies this valence diagram without claiming they do not exist or have exactly no effect on molecular energies.

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