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Calculate the number of covalent bonds in $\ce{C3O2}$.

Structure:

Lewis-like structure of C3O2

I am confused. I can count covalent bonds as 8 by counting single-single bonds, but also I can say that there are 4 "double covalent bonds". Which is correct?

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    $\begingroup$ Eight covalent bonds is correct $\endgroup$ – Maurice Sep 16 at 12:16
  • $\begingroup$ @Maurice Any Sources from where I can confirm? $\endgroup$ – Rasputin Sep 16 at 12:40
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    $\begingroup$ Four double bonds is equivalent to eight covalent bonds. $\endgroup$ – Maurice Sep 16 at 14:38
  • $\begingroup$ @Maurice But “Double Covalent Bond Is A Type Of Covalent Bond” so probably could be counted as 4 too . $\endgroup$ – Rasputin Sep 16 at 14:42
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    $\begingroup$ @Maurice The physical reality is that there are 4 bonds. The character of those bonds is partly theoretical. The term "covalent bonds" does not specifically or uniquely apply to just single bonds, at least in most terminology that is widely used. $\endgroup$ – matt_black Sep 16 at 15:20
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Widely used terminology might not match what an examination wants

It is possible that your question wants the answer 8 in this context, but whether this answer would be widely accepted by anyone familiar with bonding theory is doubtful.

In the early days of theories about bonding Lewis described bonds as consisting of two shared electrons. You are probably familiar with "Lewis Structures" which are simple pictures of bonding where electrons are represented with two dots between atoms. This is also how the theory of bonding is initially introduced to students. In this view the number of bonds in the molecule is, indeed, 8. But this–vastly over simplistic theory–is not good for explaining the bonds in many real compounds. Hence the dispute about the correct terminology.

A good theory of bonding needs to take quantum mechanics (and the shape of electron orbitals) into account and those were not understood in the early 20th century when Lewis developed his explanation of bonding.

But we don't need any quantum obfuscation to see why Lewis theory doesn't work well and why it is better not to pretend that bonds consist of pairs of electrons. Take benzene as an example. You could claim that the bonds in benzene contain 18 electrons so make up 9 bonds. But this means alternating double and single bonds. But that can't be right as all the bonds in benzene are the same length. It is better to say than benzene consists of 6 bonds each of which have somewhere between the character of a single and double bond.

This inability of the Lewis model of bonding to explain real world structures is very common. Nitrate ions are one example where all the bonds have characteristics of more than a "single" bond but less than a "double" bond. Many boranes have bonds between 3 atoms but only two electrons to share among them.

So equating bonds to shared pairs of electrons is not helpful. Nor is using terminology that defines a bond that way. So real chemists should be reluctant to count bonds that way. Whether this is the answer your question is looking for is less clear as you might still be at the stage where they have taught you Lewis structures and nothing else.

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  • $\begingroup$ I actually am taught Lewis Structure , VSEPR, And Molecular Orbital Theory ( using the LCAO method)..(A bit simplistic approach valid for less than equal to 20 electron species)...I know about resonance and the case you mentioned and I agree that the examination does require “8” to be the answer....I wouldn’t count bonds in this way at all but the examination wants what it wants...I just wished to clear up the confusion! $\endgroup$ – Rasputin Sep 16 at 18:17
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    $\begingroup$ @ Matt black. Congratulations for your clear and clever development. I am not joking ! It is nice and bright. $\endgroup$ – Maurice Sep 16 at 19:35
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    $\begingroup$ That is a really good answer! Kudos. I would like to bounty it, but that is not yet possible; I'd appreciate a reminder. To the matter at hand: carbon suboxide is probably an equally terrible example to count bonds as for the case with benzene. It has a very shallow bending potential, which indicates a much more complex electronic structure than what is possible to explain with Lewis-like structures. And as far as I remember there is/was some debate from different points of view. Certainly interesting, certainly too much for the undergrad level. $\endgroup$ – Martin - マーチン Sep 17 at 12:50

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