I guess if we look at this problem from a "conjugate" perspective then the conjugate of a covalent bond is two elements with electrons lying around. On the other hand, with two electrostatically attracted things, the lack of a bond leaves two separate, charged particles.

But that still doesn't answer the question; why are electrons less stable than ions? I guess in the case of an ion though there are still electrons but also a nucleus the stabilize the electrons.

  • $\begingroup$ What makes you think ionic bonds are weaker than covalent bonds? It seems to me that ionic bonds are stronger than covalent bonds. $\endgroup$
    – Papul
    Dec 16, 2014 at 16:50
  • $\begingroup$ think about this question... does it take more energy to separate ionic compounds into their non ionic elemental state or does it take more energy to separate two covalent bonded atoms into their non ionic elemental state... I know apples and oranges ... NaCl vs H2O? $\endgroup$
    – sergio
    Sep 13, 2017 at 19:54
  • $\begingroup$ But can two ions come together to form a covalent bond? In that case, the "conjugate" is no longer two neutral atoms. $\endgroup$ Sep 20, 2017 at 23:56

4 Answers 4


Ionic bonds can easily be as strong as covalent ones.

First off, let's be clear that almost everything has some ionic and some covalent character. Moreover, it's not an either-or situation. Some bonds have both strong ionic and strong covalent character.

For now, I'll stick to simple cases.

Let's consider $\ce{LiF}$, which seems a prototypical ionic bond. The interatomic distance is $\pu{156 pm}$ according to NIST. That gives a potential energy of $\pu{903 kJ/mol}$. Now, that assumes a full electron is transferred from $\ce{Li+}$ to $\ce{F-}$, which isn't the case (more like $0.8 - 0.9$). This simple calculation also ignores Born repulsion which is similar to the short-range Van der Waals repulsion between atoms.

A more complete analysis J. Chem. Educ. 2000, 77 (8), p 1076 gives a much better agreement. The real bond dissociation energy is $\pu{577 kJ/mol}$ (Source, PDF via the Internet Archive). A simple spherical ion model from the paper gives the bond enthalpy as $\pu{608 kJ/mol}$.

calculations based on a simple ionic model reproduce the bond energies of $\ce{LiF(g)}$ and $\pu{NaF(g)}$ to within $5\%$.

If we look up covalent bond dissociation energies, the triple bond in dinitrogen is $\pu{945 kJ/mol}$. Ok, that's perhaps stronger, but $\ce{LiF}$ only has a "single" charge and clearly some ionic compounds have greater charge transfer.

More "prototypical" covalent bonds include $\ce{C-H}$ ($\pu{410 kJ/mol}$) and $\ce{O=O}$ ($\pu{497 kJ/mol}$), both less than $\ce{LiF}$.

Now, as I said earlier, it's not the case of a bond being only covalent or only ionic.

To quote from R. Gillespie J. Chem. Educ. 2001, 78 (12), 1688:

The large atomic charges in $\ce{BF3}$ and $\ce{SiF4}$ are important to understanding the great strength of these bonds—which are the two strongest known single bonds, with bond dissociation enthalpies of $613$ and $\pu{565 kJ/mol}$, respectively. In comparison, the $\ce{C–C}$ bond in ethane has a bond dissociation enthalpy of only $\pu{345 kJ/mol}$. The great strength of the $\ce{BF}$ and $\ce{SiF}$ bonds arises from both the relatively large amount of shared density and the large atomic charges. So it is not strictly correct to say that these bonds are “predominately ionic”; if we are forced to use these terms we have to say that they are both strongly covalent and strongly ionic.

Finally, you can of course treat both covalent and ionic bonds using MO theory. For example, an approximate picture of $\ce{LiF}$ under MO theory might be something like this:

Hypothetical MO diagram of LiF

We readily see $\ce{LiF}$ has very high ionic character. For all intents and purposes, the "bonding" $\sigma$ orbital is localized on the $\ce{F}$ atom, there's very little energy difference between the $\pi$ and $\sigma$ here.

  • $\begingroup$ Your numbers are wrong for LiF, It should be 577KJ/mol. see: nist.gov/data/nsrds/NSRDS-NBS31.pdf $\endgroup$
    – Dale
    Dec 18, 2014 at 14:15
  • $\begingroup$ I like the care you have taken in choosing the examples in your answer, but I am also a little concerned that you are simply calculating the potential (I assume you just calculated the PE for separation of two charges from the distance) for LiF, while you used actual bond dissociation energies for $\ce{N2}$. That does not appear to be a fair comparison. However, I will change my answer so that it uses LiF as well as I think it is a better example to use. $\endgroup$
    – selkie222
    Dec 18, 2014 at 15:53
  • 1
    $\begingroup$ How do you reconcile these two statements: "they are both strongly covalent and strongly ionic" "We readily see LiF has very high ionic character." You don't seem to stick to simple cases. $\endgroup$
    – Dale
    Dec 19, 2014 at 22:56
  • 1
    $\begingroup$ How does BF3 have strong ionic and covalent character? Aren't shared e- density and highly localized charges mutually exclusive? Or are you saying that there is a lot of shared e- density between B and each of the Fs but also B and each F takes on a significant degree of partial charge? $\endgroup$
    – Dissenter
    Dec 19, 2014 at 22:58
  • $\begingroup$ The article by Gillespie argues you should avoid the phrase "ionic character" because "ionic" and "covalent" aren't mutually exclusive. It's pretty clear from dipole moment in diatomic LiF and quantum calculations that ~0.9 electrons are transferred. OTOH, B-F and Si-F clearly have both shared electron density AND electrostatic contributions, so both are important in the bonding in these compounds. $\endgroup$ Dec 19, 2014 at 22:59

Using your "conjugate" way of seeing two covalently bonded atoms as electrons, let's try $\ce{O_2}$ which gives you "$\ce{2O^{\bullet}}$". If you think about how the electrons are arranged in terms of electron configuration, you will see that $\ce{O^{\bullet}}$ has $\ce{[He]}2s^2 2p^4$ which is not a particularly stable electron configuration (if you only know the "secondary school electron configuration" you should recognise this as 2,6). Not having a full electron shell (2,8) means this is unstable.

If you do the same with $\ce{NaCl}$, you get $\ce{Na+}$ and $\ce{Cl-}$. Both of these have full shell electron configurations ($\ce{Na+}$ is $\ce{[Ne]}$ or 2,8; $\ce{Cl-}$ is $\ce{[Ar]}$ or 2,8,8). So the ions are indeed more stable in this sense.

However, relative stability of ionic and covalent bonding cannot be simply compared using your "conjugate" method, because you need to make sure that you are making the comparison under similar conditions. For example, you cannot really say $\ce{NaCl}$ is less stable than $\ce{O2}$ because NaCl dissociates into ions in water while $\ce{O2}$ does not, because the surrounding water molecules have a very strong dipole which will interfere much more strongly with ionic bonding (which is electrostatic) than covalent bonding in a homonuclear diatomic (which has no reason to dissociate into neutral radicals in water). You can possibly compare hardness of $\ce{NaCl}$ versus diamond, i.e. carbon that is purely covalently bonded, but graphite would not make such a good comparison because between the layers of graphite there are only weak Van der Waals interactions (and it is therefore softer because you are overcoming VdW interactions not covalent bonds). There are also special cases such as ionic salts with a particularly low melting point, and there are many examples of weak covalent bonds. You also want to pick examples with few confounding factors, such as large differences in electronegativity which can give ionic contributions to covalent bonds, or large differences in size and effective charge which can give covalent contributions to ionic bonding.

I will answer your question using two examples of giant lattices that show near-ideal ionic and covalent bonding: $\ce{LiF}$ versus diamond (with acknowledgements to Geoff Hutchison for the choice of LiF). Lithium, fluorine and carbon are all in period 2, so avoiding further complications due to differences in size such as comparing to $\ce{NaCl}$. I am "taking a shortcut" here by comparing boiling points, which is a bulk property that you can use for a giant lattice, because it is difficult to find a pure ionic diatomic.

$\ce{LiF}$ has a boiling point of 1909 K, while diamond sublimes at 3900 K. By this measure, it would appear that covalent bonding is stronger than ionic bonding in this case.

In $\ce{LiF}$, the distance between the lithium and fluoride ions is 202 pm*, while in diamond the distance between carbon atoms is 154 pm. In $\ce{LiF}$, the two oppositely charged ions are further away than the (positively charged) nuclei of any two carbon atoms in diamond that is held together by (negatively charged) electrons between them. What I am trying to point out is that, both forms of bonding involve interactions of positive and negative charges, but in covalent bonding the charges are much closer together, so the attractive electrostatic forces are much greater. This is why in this particular comparison, diamond comes out with stronger covalent bonding than the ionic bonding found in $\ce{LiF}$.

As I indicated in the third paragraph, there are many cases where the bonding within an ionic substance can be stronger or weaker compared to a covalent substance. This is why I have emphasised that everything that is discussed is only true for this particular comparison. It will be unwise to claim that one type of bonding is overwhelmingly stronger than another, because it always depends on what exactly is being compared. Given the near infinite combinations of atoms that can give rise to bonding that is ionic, covalent, and a mixture of the two, it would be difficult to make a full catalogue of covalent and ionic substances and compare these against each other.

*computed by taking half the distance of the cubic lattice.

  • $\begingroup$ "It will be unwise to claim that one type of bonding is overwhelmingly stronger than another". Definitely agree! $\endgroup$ Dec 18, 2014 at 19:09

Short answer:

To quote Mehrdad: You're basically comparing an intermolecular force with an intramolecular force.

Ionic bonds are two charged atoms sitting next to each other with minimal overlap of their orbitals.

Covalents bonds are two atoms becoming one molecule with significant changes in orbital structure.

It is not always clear cut: there are atoms with part-ionic and part-covalent character or two atoms sitting next to each other with moderate overlap of their orbitals.

Here is a good explanation (for solids): Are metallic/ionic bonds weaker than covalent bonds?

Long answer:

Electrostatic bonds by definition are incapable of any significant bond formation. Bond formation involves the formation of a lower energy "bonding orbital(s)" out of two (or more) higher energy ordinary orbitals on adjacent atoms within the molecule. The formation of bonding orbitals gives covalent bonding orbitals much lower energy (not smaller, see diagram) than non-bonding orbitals which exist in the unbound state. This energy difference is larger than is possible for merely electrostatic attractions. Why is this the case? Because the electrons in bonding orbital are better able to counteract the positive charges of the adjacent nuclei. This makes sense because the negative charge on one molecule in an electrostatic arrangement is only able to neutralize the positive charge from its own nuclei. Bonding orbitals that are only present in covalent bonds share those electrons between two nuclei and thereby better neutralize the charges of both nuclei, which results in a lower energy for the bonded system than is possible in an ionic system.

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  • $\begingroup$ What's inherently better about bonding orbitals vs electrostatic attraction? $\endgroup$
    – Dissenter
    Dec 17, 2014 at 4:30
  • $\begingroup$ So...stabilizing electrons across multiple nuclei vs positive attracts negative ... What's the difference? $\endgroup$
    – Dissenter
    Dec 17, 2014 at 4:43
  • $\begingroup$ -1. The electrostatic attraction is small? Calculate the Coulomb interaction in $\ce{LiF}$ (ionic, + small nuclei) or $\ce{Al^{+2}O^{-3}}$ as part of $\ce{Al2O3}$. $\endgroup$ Dec 17, 2014 at 5:12
  • $\begingroup$ @GeoffHutchison Al2O3 has a mixture of ionic and covalent character. Without the covalent character the bonds would be much weaker to the point of being non-competitive with covalent bonds. The same goes for LiF. $\endgroup$
    – Dale
    Dec 18, 2014 at 14:34
  • $\begingroup$ Everything has some covalent character, but I think it's silly to insist the covalent contribution is so high. I'll find numbers later. $\endgroup$ Dec 18, 2014 at 16:08

It really depends on which covalent bonds and ionic bonds we are comparing.

Generally, the strength of a bond, ionic or covalent, depends on two things:

  • The atomic radius of the elements which take part into the bond
  • the difference in electronegativity between those 2 elements
  • 1
    $\begingroup$ I think this answer suffers from the fallacies of the question. If not you should explain each of the factors you mentioned. $\endgroup$
    – M.A.R.
    Jun 6, 2019 at 15:16
  • $\begingroup$ There is nothing to be explained.Those 2 things come from theory.What else I forgot to mention is how many chemical bonds the elements form and if there are any bonds where the bonded pair comes from the same orbital which is of course an anti bond. $\endgroup$ Jun 6, 2019 at 16:03
  • $\begingroup$ I mean where there are 2 bonded pairs and their electrons come from the same orbital for example an atom uses 2 electrons from its 1s orbital and the bonded atom uses 2 electrons from its 1s orbital. $\endgroup$ Jun 6, 2019 at 16:12

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