-1
$\begingroup$

A level chemistry student here, fairly basic question. We always learnt (from GCSE bond enthalpies) that "Bigger bonds are stronger", so to speak; my teacher often quoted $\ce{N#N}$ as very unreactive as the triple bond is very strong. However, now we are covering organic chemistry in a bit more depth at A-level, they are saying that the alkanes are generally unreactive as the $\sigma$-bond (which I read as: single bond) is fairly strong, but the $\pi$-bond (read: double bond) has more exposed electrons and so breaks more easily.

This is quite hard to accept, considering it contravenes everything I've ever learnt about bonds (more shared electrons will surely exert a stronger electrostatic attraction and have a higher bond enthalpy, no?) but then again we are not taught about bond energies in any great detail. I am hoping someone could explain or motivate this a little (I imagine a fully correct answer would be way over my head ;)): why are double/triple bonds sometimes weaker, but most of the time stronger, than single bonds? And what properties of the different elements (e.g. this holds for carbon but not for nitrogen) cause the discrepancy?

chem.libretexts.org Bond_Energies:

Bond Bond energy [kJ/mol] Incremental bond energy [kJ/mol]
$\ce{C-C}$ 347 -
$\ce{C=C}$ 614 264
$\ce{C#C}$ 839 225
$\ce{N-N}$ 160 -
$\ce{N=N}$ 418 258
$\ce{N#N}$ 941 523
Improve this question
$\endgroup$
8
  • 2
    $\begingroup$ Your teacher told you that nitrogen gas is a powerful explosive...? $\endgroup$
    – orthocresol
    Commented Apr 6, 2022 at 12:38
  • 1
    $\begingroup$ Well, that the $\ce{N#N}$ bond is why nitrogen features in many explosives (TNT I think? Things like that) @orthocresol because when it is broken, it releases an unusually large amount of energy as its bond enthalpy is quite high (so we are told) $\endgroup$
    – FShrike
    Commented Apr 6, 2022 at 12:40
  • 8
    $\begingroup$ We are all doomed as atmosphere is full of $\ce{N#N}$ !! :-) // No, TNT does not contain $\ce{N#N}$. //It is exactly the opposite, you need enermous amount of energy to break the $\ce{N#N}$ bond, that is why $\ce{N2}$ is so little reactive, not reacting directly with most of elements. $\endgroup$
    – Poutnik
    Commented Apr 6, 2022 at 12:40
  • 1
    $\begingroup$ :) Oh ok in the converse then - I must have misremembered - $\ce{N#N}$ is very unreactive as the bond enthalpy is so high @Poutnik $\endgroup$
    – FShrike
    Commented Apr 6, 2022 at 12:41
  • 1
    $\begingroup$ But, the question stands (I'll edit out the explosive part) $\endgroup$
    – FShrike
    Commented Apr 6, 2022 at 12:41

1 Answer 1

Reset to default
3
$\begingroup$

The contexts are different here. Yes, nitrogen features in many explosive compounds because the explosion forms nitrogen gas. In this case, the N–N triple bond does not even exist in the starting material, so 'all three' of the bonds have to be formed at once, and that releases that huge amount of energy (941 kJ/mol according to your table).

Conversely, when one talks about alkenes or alkynes being relatively reactive, the issue is not about how strong the triple bond itself is. These reactions usually involve a double bond being converted to a single bond, or a triple being converted to a double. Thus, the energetic term under consideration is only the difference between the double and a single bond, which is much smaller.

So, it's not so much that nitrogen is special and carbon isn't. It's true that the nitrogen bonds are slightly stronger than the carbon bonds, but that is not really a factor here. The main issue is that you're comparing two different things (formation of a full triple bond vs the breaking of 'half' a double bond, or perhaps even less than half).

Improve this answer
$\endgroup$
3
  • $\begingroup$ Right, and the difference between a $\pi$ and $\sigma$ bond's energy is less than the energy it takes to fully break a $\sigma$ bond, but if you were to attempt breaking both at the same time then the double bond is still stronger. Have I interpreted correctly? $\endgroup$
    – FShrike
    Commented Apr 6, 2022 at 12:57
  • 2
    $\begingroup$ Of course. Such reactions generally don't happen, and if they do, they happen stepwise in that the double bond is first converted to a single bond which is then broken. $\endgroup$
    – orthocresol
    Commented Apr 6, 2022 at 12:58
  • 1
    $\begingroup$ @FShrike Yeah, seems like we all would like for your question to be better than that, but it seems to boil down to that a "breaking" of double bond only reduces it's order to one and new single bonds are introduced. That makes the title of question misleading and then post should just been closed as unclear. $\endgroup$
    – Mithoron
    Commented Apr 6, 2022 at 13:08

Not the answer you're looking for? Browse other questions tagged or ask your own question.