A question about ionic bonding

[resurgance2001]

I am sure this question will have been asked elsewhere, so please forgive me if it has.

My question involves the transfer of electrons in ionic bonding. We are told that a metal will give up its outer most electron because by doing so it will gain the electron configuration of a noble gas which is stable. This is the simple explanation which is given at GCSE and I think A level also. However, my question is really, why? It still requires energy to remove the electron (1st ionisation energy) and it further requires energy (no?) to force an additonal electron to be accepted by a halogen for example. So where is this energy coming from? Does the 'stability' of a full outer shell involve energy in some way? Thanks

 

[DrDu]

The point is that electrons in the same shell are not very efficient in screening each other from the charge of the nucleus. For an alkali metal, this is of little relevance as it only has one electron in the valence shell, so it will always see about 1 nuclear charge. On the other extreme in halogens, an electron feels attracted by more than 1 nuclear charge unit. Hence electrons in non-metals are much more strongly bound than in metals.

 

[mjc123]

You have to consider all the energies involved in the reaction of (say) Na(s) and Cl₂(g) to give NaCl(s). Have you come across the Born-Haber cycle?

 

[resurgance2001]

Yes - I have done questions about the Born-Haber cycle and I do appreciate the arguments about screening.

I still struggling though to understand from the point of view of energy why the electron is actually transferred say from the sodium atom to the chlorine atom. Because what we are 'told' (apart form the screening argument) is that the atom 'wants' to gain the electron configuration of a noble gas. This seems to be implying that the stability of a noble gas configuration by itself is a driving force the reaction to take place. This actually came from a student of mine who, having heard the standard arguments that are given in all the textbooks, just said she wasn't satisfied and didn't get it. Then I realized that I had not really thought about it myself and had just accepted without questioning the explanation that atoms 'want' to gain a noble gas configuration. Is there, for example, some intrinsic lowering of potential energy when the out shell is full? Or does entropy come into play here is some way. Or is it (I suspect it might be) some quantum mechanical effect at play here. What is so special about having a full outer shell? There must be energy involved in this somehow.

 

[Ygggdrasil]

it further requires energy (no?) to force an additonal electron to be accepted by a halogen for example.

No, the halogens and most other non-metals have positive electron affinities, meaning they release energy when they accept an additional electron: https://en.wikipedia.org/wiki/Electron_affinity

The electron affinity of the halogens is comparable to the first ionization energy of sodium and the other alkali metals.

 

[resurgance2001]

Ok - that makes more sense now. Thanks

 

[DrDu]

No, the noble gas configuration is not an attraction point per se. The point is rather that both electron affinity and ionisation energy show a saw tooth like dependence when plotted as a function of the number of the number of electrons. Ionisation energy is lowest for a valence shell containing just one electron and highest for a noble gas shell, while electron affinity is highest for a noble gas and lowest for an alkali metal. Hence an atom with less than a half filled shell can continuously reduce energy by giving electrons to atoms with a more than half filled shell. The process comes to an end when the valence shell is empty, as further ionisation would require to remove electrons from a completely filled shell, which requires much more energy. On the other hand, an atom with more than half filled shell can take up further electrons with only a modest increase of energy as due to the reduced screening, the additional electrons will not see a neutral atom but some unscreened nuclear charge. Again, this will come to an end when a shell is completely full and a electrons would have to enter a new shell.

 

[mjc123]

See here for example: http://www.chemicalforums.com/index.php?topic=77729.msg283639#msg283639

 

[Lord Jestocost]

Ok - that makes more sense now. Thanks

Here are some data: http://hyperphysics.phy-astr.gsu.edu/hbase/molecule/Paulirep.html#c1