= Solution
\b[The elementary valence-MO model gives $\boxed{\text{bond order}(\mathrm C_2)=2}$.] Each carbon supplies four $n=2$ valence electrons, so fill eight electrons in the ordering appropriate to the lighter second-row diatomics:
$$
(\sigma_g2s)^2(\sigma_u^*2s)^2(\pi_u2p)^4(\sigma_g2p)^0(\pi_g^*2p)^0(\sigma_u^*2p)^0.
$$
The two $\pi_u$ orbitals are degenerate, as are the two $\pi_g^*$ orbitals. The $2s$ 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 $\pi_u$ pair below the empty p-derived $\sigma_g$ orbital in this diagram. All occupied orbitals are paired.
\Image[/past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1-carbon-orbitals.png]
{title=Original valence <molecular orbital> energy diagram for carbon dimer, with atomic inputs and symmetry labels}
The omitted $1s$ orbitals form filled $\sigma_g1s$ and $\sigma_u^*1s$ 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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