Which one of the three compounds has the highest $\ce{C-C}$ bond stability?

Ethane has $\mathrm{sp^3}$ hybridisation so less $\mathrm{s}$-character and lesser electronegativity compared to ethyne. Ethyne has high $\mathrm{s}$-character and high electronegativity. Therefore the covalent character of $\mathrm{sp^3-sp^3}$ in Ethyne should be stronger and have more stability right? the order would therefore be $\mathrm{sp-sp}$ > $\mathrm{sp^2-sp^2}$ > $\mathrm{sp^3-sp^3}$

Some online sites tell so, but others say since the $\ce{C}$ atoms are so close in ethyne that the nuclear repulsion would make it less stable. So ethane's $\ce{C-C}$ would be stronger and that therefore the order should be reversed

Edit: nutshell- sigma bond of which hybridisation is strongest among $\mathrm{sp}$, $\mathrm{sp^2}$ and $\mathrm{sp^3}$? I took ethane, ethene, and ethyne just as a example

  • $\begingroup$ "Stability" is another word for "energy difference between one state and another". Now which are those for each of your three suspects? $\endgroup$
    – Karl
    Commented Oct 12, 2020 at 20:49
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    $\begingroup$ Another way to look at this problem is to compare the bond dissociation energies. For C-C, it is about $350$ kJ/mol. For C=C in ethene, it is $610$ kJ/mol. And for ethyne, it is $835$ kJ/mol. $\endgroup$
    – Maurice
    Commented Oct 13, 2020 at 20:42
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    $\begingroup$ Well one can argue about the single sigma bond in the three cases. This reflects on the stability of the whole skeleton. It cannot be accessed experimentally but it is a legit question. Especially writing out C-C. I don't know what OP is asking, but I guess this. Otherwise the answer is really evident. @Mathew Mahindaratne and at Maurice $\endgroup$
    – Alchimista
    Commented Oct 14, 2020 at 13:33
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    $\begingroup$ @Maurice well that's what i found, but isn't that including the pi bonds too in the C−C bond? i wanted to know specifically for the sigma bond, whether there's any difference in stability between a sigma bond of sp, sp2, sp3 $\endgroup$ Commented Oct 15, 2020 at 12:09
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    $\begingroup$ "Stability" is a dodgy term that means very different things in different contexts. So which version are you looking for here? If you mean bond strength, say so. If you mean ability to undergo different reactions, say so (systems with multiple bonds can be far more reactive in certain reactions even though the overall bond is stronger). $\endgroup$
    – matt_black
    Commented Oct 16, 2020 at 11:26

1 Answer 1


The truth is: in reality, sigma bonds, bond orders and hybridization do not exist (not even bonds or orbitals do). They are just concepts to explain the reactivity, stability and geometry of a molecule (and in the case of orbitals to build approximate many-particle wave functions). There is thus no way to compare a sigma-bond strength without choosing some model which introduces bias.

The least-biased way to get there is probably starting from the experimental dissociation energy already mentioned. These can be divided by the bond order (which, again, is not a thing, but quite well established):

  • ethane: 350 kJ/mol -> 350 kJ/mol per bond
  • ethene: 610 kJ/mol -> 305 kJ/mol per bond
  • ethyne: 835 kJ/mol -> 278 kJ/mol per bond

In general: for any chosen approach, the sigma bond in ethane will be the strongest even though the bond distance of ethyne is the smallest. The additional electrons of the pi bond(s) destabilize the bond.

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    $\begingroup$ I do not think it is particularly helpful to point out how everything does not exist. How do you define bond dissociation energy if you have already claimed that bonds do not exist? $\endgroup$
    – user85426
    Commented Oct 16, 2020 at 8:22
  • $\begingroup$ A dissociation energy is a thing, it is defined as E(Fragments) - E(Molecule) without any assumption on bonds. The reason I point out, that all of this does not exist, is that chemists (like me) tend to forget that. Students often take these concepts literally. $\endgroup$
    – Libavi
    Commented Oct 16, 2020 at 8:27
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    $\begingroup$ I think this answer touches on some philosophical and semantic issues. First, I don't see how what you note is sad. It's just facts. What might make it sad is how popular some of these models are and how many of them are abused. Moving on. I don't think it is wise to claim (for all intents and purposes) that bonds do not exist. The electron density is observable, as are nuclei, we can measure the energy it takes to change one molecular structure to another. Using the term bond just makes things simpler. Lastly: Geometry, that's such a fuzzy term for structure (imo). $\endgroup$ Commented Oct 16, 2020 at 10:30
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    $\begingroup$ Libavius, I did down vote because you enter into much philosophy. You can surely saturate carbon atoms in steps and this is an evident bond order. As it seems that you are aware of the sophistication implicit in these treatments, it should also be that you recognise the validity and depth of the question. Equally dividing the energy per bond just tell us that some bonds are less stronger. I can say that the bond in Ethyne is the stronger because is still there in spite of destabilising extra electrons. Still, the idea is good. Like in benzene, for which delocalisation does not come in play bec $\endgroup$
    – Alchimista
    Commented Oct 17, 2020 at 17:57
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    $\begingroup$ .. Because of an aromatic sextet, but rather the latter is forced to delocalize by the strength of the sigma skeleton. This said I am not sure of which sigma bond is stronger, but I am pretty sure applying theory provides the result. The fact that it can be void of practical importance is clear to me, too. $\endgroup$
    – Alchimista
    Commented Oct 17, 2020 at 18:00

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