In an ionic bond, are electrons actually transferred from one atom to the other to form a positive and a negative ion OR are ionic bonds just extremely polar covalent bonds? Sorry if this seems like a simple question but I've heard this somewhere so just wanted to make sure.

  • $\begingroup$ Have a look at this answer of mine. It is quite related to your question. $\endgroup$ – Philipp Oct 16 '14 at 21:04
  • $\begingroup$ So are you saying the electrons in say Cl in NaCl are just very far from the nucleus (so it isn't in Na's shells)? $\endgroup$ – user58953 Oct 17 '14 at 6:02
  • $\begingroup$ I wouldn't put it that way. I think you termed it quite good with "extremely polar covalent bonds". You usually find some covalent contribution left in every bond. It might be very, very weak but it should be there. However, describing some bonds such as that in $\ce{NaCl}$ as purely ionic has advantages in most situations because it makes the treatment easier while giving quite accurate results. $\endgroup$ – Philipp Oct 17 '14 at 7:24

This is not a simple question at all, in fact, it is an interesting insight in chemical bonding. You make a valid assertion with your question:

are ionic bonds just extremely polar covalent bonds?

This answer is a simplistic way of looking at it.

According to ChemGuide page Electronegativity, there is no real clear-cut 'boundary' between covalent and ionic bonding. The website describes that there is a 'spectrum of bonding' based on the electronegativity (chart shown below) of the constituent elements.

enter image description here

Image source: UC Davis ChemWiki

So, the greater the electronegativity, the greater the polarity or "pulling power" on the bonding electrons towards elements with a higher electrongativity.

In the following diagrams (from the ChemGuide site linked before), A and B represent arbitrary elements, and the green and red dots represent bonding electrons. Starting from a situation where the polarity is zero as the electronegativity of the constituent elements are the same (e.g. diatomic molecules $\ce{O2}$, $\ce{Cl2}$ etc), as in diagram I.

As the polarity increases, the bonding electrons move towards the more electronegative member, causing the bond to become increasingly polar (II), then eventually to the point where the difference is enough (III) for:

To all intents and purposes, A has lost control of its electron, and B has complete control over both electrons. Ions have been formed.

I bolded the 'to all intents and purposes' to emphasise that this is a simplistic generalisation

enter image description here

enter image description here

enter image description here

As the lowest electronegativity is not zero (it is Francium at 0.7), there will always be an influence on the bonding electrons. So, the reality is that A does not completely lose its influence on the bonding electrons.

For example, if we were to look at a compound combining the most and least* electrongative elements (as appear in the diagram above): francium fluoride $\ce{FrF}$, even though fluorine has a far stronger electronegativity than francium, the electronegativity of the latter is not zero, so, the influence of the francium is very small, but not zero - indicating that purely ionic bonds don't exist.

  • As noted in the comments, cesium is likely to be less electronegative than francium, however the point made above still stands.

This is nicely summarised by an article from the University of Malta:

Pure ionic bonding is not known to exist. All ionic compounds have a degree of covalent bonding. The larger the difference in electronegativity between two atoms, the more ionic the bond.

  • $\begingroup$ Francium is probably more electronegative than caesium due to relativistic effects. $\endgroup$ – Mithoron Jun 25 '15 at 10:52
  • $\begingroup$ True, but it still makes a good example. Asides, I was just using a simplistic example - I'll make that more explicit. $\endgroup$ – user15489 Jun 25 '15 at 19:33
  • $\begingroup$ As a side question: has FrF ever been synthesized, as far as you know? $\endgroup$ – Enrico Maria De Angelis Jul 2 '17 at 17:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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