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Is bond formation "strictly" exothermic? The IUPAC definition of exothermic doesn't make any reference to bond formation. However, I have seen the aforementioned statement before - that bond formation is "strictly" exothermic.

I suspect that the answer is "it depends" and that "it depends" at least on how accurately bonds are characterized. For example, the first reaction below is endothermic, while the second one is exothermic:

$\ce{2NO -> ONNO}$.

$\ce{2NO_2 -> O_2NNO_2}$.

Lewis structure analysis for the first reaction at best misleads because one might suspect the nitric oxide molecule as having a bond order of 2.

enter image description here

However, the lone electron is actually delocalized. So the actual $\ce{N-O}$ bond order should be 2.5.

enter image description here

In the $\ce{ONNO}$ molecule however there is no delocalization of electrons and the bond order of the $\ce{N-O}$ bond is 2. So the $\ce{N-O}$ bond is weakened, and even the formation of the $\ce{N-N}$ bond cannot compensate for this weakening.

Also, the above example reminds me of certain unstable homonuclear species. Such as $\ce{Ne_2}$, which according to MO theory has a bond order of 0 - i.e. there is no bond. So I suppose that bond formation is not always exothermic. On the other hand, if the bond order is 0, is a bond really "formed"?

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4 Answers 4

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Bond formation is alway strictly exothermic in the sense of the change of enthalpy.

exothermic reaction A reaction for which the overall standard enthalpy change $\Delta H^\circ$ is negative.

A bond can only exist, if it needs energy to break it, i.e. the bond dissociation energy is always positive.

bond-dissociation energy, $D$ The enthalpy (per mole) required to break a given bond of some specific molecular entity by homolysis, e.g. for $\ce{CH4 -> .CH3 + H.}$, symbolized as $D(\ce{CH3−H})$ (cf. heterolytic bond dissociation energy).

This has absolutely nothing to do with a reaction being exothermic/endothermic or exergonic/endergonic, because this is defined by the rearrangements of bonds.


Regarding noble gas diatomics, it is quite clear from MO-Theory, that there is no bond. However, even these non bonded elements have a non-zero dissociation energy. Please refer to "Why are noble gases stable" and to answers and comments within.

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  • $\begingroup$ +1 for bringing up the need to consider bond breaking as well as bond breaking when determining whether or not a reaction is exothermic. $\endgroup$ Jun 13, 2014 at 10:26
  • $\begingroup$ What do you mean noble gases have a non-zero dissociation energy? How can non-bonded elements have "dissociation" energies? $\endgroup$
    – Dissenter
    Jun 13, 2014 at 14:08
  • $\begingroup$ Admittedly they are very, very small, but not negligible. There are weaker forces at play, i.e. London Dispersion and van der Waals. These are also part of the reason, why noble gases can be liquid close to zero Kelvin. $\endgroup$ Jun 13, 2014 at 15:01
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    $\begingroup$ Martin, it does have to do with a reaction being exothermic or endothermic, because you need to define another state that you are comparing to. Here you are comparing methane to methyl radical and hydrogen radical, which is a specific reaction. Weak bonds do form between noble gas atoms: scitation.aip.org/content/aip/journal/jcp/98/4/10.1063/1.464079 sciencedirect.com/science/article/pii/S0009261401010880# What about the bonds of trigonal bipyramidal transition states? en.wikipedia.org/wiki/Transition_state Do they require energy to break? $\endgroup$
    – DavePhD
    Jun 13, 2014 at 15:16
  • $\begingroup$ @Dave I am not comparing anything with anything else, I am just quoting the IUPAC Goldbook. This definition is complete in a concise and theoretical way. And with this given, the BDE is always positive. Experimentally you do not need to actually break the bond, you can derive it from spectroscopy. Thank you for the literature. As for the 'bonds' in the transition states, they do not fit within the definition of bonds or chemical bonds. $\endgroup$ Jun 13, 2014 at 16:04
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At very high pressure, say $>\pu{100 GPa}$, diatomic molecules such as $\ce{H2}$ and $\ce{O2}$ are no longer favorable relative to monoatomics, even at low temperature where Gibbs free energy $G$ is dominated by enthalpy $H$.

See Phase Diagram of Hydrogen.

For example the cores of Jupiter and Saturn are monoatomic hydrogen.

Diatomic bond formation would be endothermic under such conditions.

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  • $\begingroup$ Dave do you know why that is? I would have thought that a diatomic occupies less space than two monoatomics and so at high pressure the diatomic side of the equilibrium would be favored. $\endgroup$
    – ron
    Jun 12, 2014 at 21:27
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    $\begingroup$ The diatomic will have a certain optimal separation between the nuclei (bond length). My thinking is that at high enough pressure/density it is no long possible to have the nuclei so separated. Also, the monoatomic state is metallic so it is bonded in that sense, just not covalently bonded. This is related to my question here: chemistry.stackexchange.com/questions/10226/… $\endgroup$
    – DavePhD
    Jun 12, 2014 at 22:02
  • $\begingroup$ Why is the monoatomic state metallic? $\endgroup$
    – Dissenter
    Jun 12, 2014 at 22:06
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    $\begingroup$ Even though metallic hydrogen may be called monoatomic, it is bound to many other atoms at once. It's not that bonding becomes endothermic, but the localization of bonding (breaking the distributed metallic bonds to create localized covalent bonds) is. [I took too long to write this and it seems you've made the comment yourself already] $\endgroup$ Jun 12, 2014 at 22:06
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    $\begingroup$ So still, at the end of the day, bond formation is an exothermic process. If the reaction coordinate for bond formation is endothermic, then bond formation simply does not occur. $\endgroup$
    – ron
    Jun 12, 2014 at 23:11
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Looking back, I don't like my old answer so I'm adding a totally different answer.

A specific (theoretical) example of endothermic bond formation is described in Prediction of a Metastable Helium Compound: HHeF J. Am. Chem. Soc. 2000, 122, 6289-6290.

As reported in table 2:

The energy to dissociate HHeF to H + He + F is negative.

The energy to dissociate HNeF to H + Ne + F is negative.

The energy to dissociate HHeF to HF + He is negative.

The energy to dissociate HNeF to HF + Ne is negative.

However, HHeF is stabilized by being in a potential energy well, the activation energy to dissociate being relative high.

The authors conclude: "Remarkably, HHeF is also predicted to be a metastable species, which represents the first neutral compound containing a helium chemical bond."

The bond formation is endothermic.

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    $\begingroup$ Do the author's actually say that bond formation is endothermic? I can understand that formation of HHeF could be endothermic (just like the formation of dewar benzene from benzene is endothermic) because you break an HF bond and make weak H-He and He-F bonds. But do the author's claim that formation of the actual H-He or He-F bond is endothermic (the formation of the actual 1-4 bond in Dewar benzene is exothermic even though the reaction is endothermic). Do the author's distinguish between reaction endothermicity and bond formation endo\exothermicity? $\endgroup$
    – ron
    Dec 23, 2014 at 18:08
  • $\begingroup$ @ron the best I can do is quote: "The H + Ng + F dissociation limit is strongly dependent on the nature of the noble gas: exothermic for He and Ne, near thermoneutral for Ar and endothermic for Kr (Table 2). On the other hand, fragmentation to HF + Ng is predicted to be a strongly exothermic process for all species." (where Ng is generic for noble gas). $\endgroup$
    – DavePhD
    Dec 23, 2014 at 18:19
  • $\begingroup$ It seems to me they are discussing bond breaking (fragmentation, dissociation) not bond formation. Do you read it differently? $\endgroup$
    – ron
    Dec 23, 2014 at 18:24
  • $\begingroup$ yes it is talking about dissociation, and agreeing with Martin's criteria for whether or not bond formation is exothermic (that the dissociation energy to neutral species being positive means bond formation is exothermic), that the dissociation to H + He + F has negative dissociation energy necessitates an endothermic bond formation. $\endgroup$
    – DavePhD
    Dec 23, 2014 at 18:36
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Besides these detailed explanations a picture sometimes helps to understand bond energy.

The picture below shows what happens to the potential energy when two atoms approach one another and a bond forms. The separated atoms are at an energy of zero. The bond is formed at the minimum (negative) energy (ignoring zero point energy of the bond vibrations). The dissociation energy D$_e$ is always positive. In a complex molecule this picture should apply to each individual bond.

Note that there is no scale, the bond could be normal covalent or due to dispersion forces, it makes no difference in principle to the general shape of the potential energy. Of course, if the bond energy is small, say, comparable to $k_BT$ at room temperature, the bond would only be stable at low temperatures.pES

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