By taking the oxidation state of the terminal oxygen atoms as -2, and oxidation state of the carbon atoms adjacent to these oxygen atoms as +2, we are left with the oxidation state of the middle carbon atom as 0.

However I researched a bit into the structure of $\ce{C3O2}$ and found out that it is easy to bend (and can be non-planar?), would this affect the oxidation state of the central carbon atom or would it remain as 0.

My thought process initially:

  • The oxygen atoms being more electronegative will pull the electron clouds towards themselves leaving the adjacent carbon atoms with a positive charge.
  • I then thought that this positive charge will be distributed among the carbon atoms leaving the middle carbon atom with a non zero oxidation state. (On reading the comment by Poutnik, I think this thought is wrong since it is not taking total bond breaking operation into account)
  • 2
    $\begingroup$ Why do you think it could affect the oxidation number of the central C? Remember that ON is the formal charge, obtained after undergoing formal total bond breaking operations under specific conditions. (Elaborate your reasoning in the question.) $\endgroup$
    – Poutnik
    Jun 11, 2023 at 17:53
  • $\begingroup$ @OscarLanzi No, they aren't? It's irrelevant if bonds are single or double, as long as they are between carbons. This compound was even modelled as carbido complex of two CO molecules - what you mention, but on both sides and it's still $0$ ox. state in the middle. $\endgroup$
    – Mithoron
    Jun 11, 2023 at 22:08
  • 2
    $\begingroup$ A different molecular structure can surely lead to a different chemical environment and consequently different charges. However, if the rules of oxidation state are correctly employed, the middle carbon will always be zero. Iirc the candidate partial charge also is close to zero. And the bonding pattern fits better to C(CO)2. $\endgroup$ Jun 11, 2023 at 22:19
  • $\begingroup$ The $\ce{C(CO)2}$ complex model actually looks a lot like the combination of structures you would draw using valence bond theory. The central carbon in the complex would have its pi-symmetry orbitals filled, and then they overlap with the pi acceptors in the $\ce{CO}$ ligands so pi electrons are delocaluzed the same way they would be with the ordinary valence bond theory. Bond lengths in $\ce{C3O2}$ suggest stringer multiple bonds than those in a nonconjugated alkene or ketone. $\endgroup$ Jun 12, 2023 at 18:09

1 Answer 1


First off, carbon suboxide is much closer to being linear than formally isoelectronic species such as $\ce{N5^+}$. The energy difference between linear and bent conformations is so small it gets smeared out by quantum-mechanical zero-point energy, and chemists call the structure "quasilinear".

If we work out the effective hybridization of $s$ and $p$ orbitals as a function of bond angle, a 160° bond angle which is considered the most stable configuration (Wikipedia, citing Reference 1), gives $sp^{1.06}$ (the number is the negative of the secant of the bond angle). So even if the premise of a linear geometry is ambiguous, the $sp$-hybridized electronic structure is an accurate electronic model and on that basis, the middle carbon may be assigned oxidation state $0$ (and the other carbons $+2$).

Some additional examples of carbon(0) bonded to only two other atoms is given by degruyter.com.

Note that the middle carbon comes out with a significantly lower oxidation state than the other two. This dovetails with the middle carbon being nucleophilic (accepting protons when the carbon suboxide reacts with water or hydrogen chloride) while its neighbors are electrophilic.

Cited Reference

  1. Brown RD (1993). "Structural Information on Large Amplitude Motions". In Laane J, Dakkouri M, Veken Bv, et al. (eds.). Structures and Conformations of Non-Rigid Molecules. NATO ASI Series. Vol. 410. Springer Netherlands. pp. 99–112. doi:10.1007/978-94-011-2074-6_5. ISBN 9789401049207.

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