QUESTIONS AND COMMENTS

Atomic and ionic radii compared with ionisation energies for the first transition series

This question is a good example of a conscientious student trying to read more into a syllabus statement than is really there. It came from a CIE (Cambridge International) A level student who had thought more carefully about a particular topic than was good for him!

He wanted to know: "Zinc's atomic radius is 0.137nm while copper's is 0.128 nm (taken from my A level text). So why . . . is the ionization energy of zinc higher?"

He also asked: "Like atomic radius and ionization energy, does ionic radius (for some particular charge, say 2+) follow the same pattern as atomic radius?"

The statement in the transition elements part of that syllabus said: "Candidates should be able to: . . . state that the atomic radii, ionic radii and first ionisation energies of the transition elements are relatively invariant".

What this means is that the atomic and ionic radii and first ionisation energies don't change much across a transition series. Notice that it also says "state" and not "explain".

Anyway, this particular student had looked in some detail at the data - often a dangerous thing to do at this level in chemistry! He was trying to tie together the explanations for the trends in atomic radii and ionisation energy as you go across the first transition series from scandium to zinc.

Before you read on, it will save me having to repeat stuff from elsewhere on the site if you first read the bits about transition metals on the pages about

• ionisation energies

• atomic radii

Taking the two bits of the question separately:

Zinc's atomic radius is 0.137nm while copper's is 0.128 nm (taken from my A level text). So why . . . is the ionization energy of zinc higher?

The first ionisation energy of copper is +746 kJ mol^-1; zinc's is = +906. (These values vary slightly depending on what data source you use, but only by a kJ or two.) The explanation for the higher zinc value is fairly straightforward and you will find it on the ionisation energy page above.

A high ionisation energy is produced by factors which cause the outer electrons to be more strongly attracted to the nucleus. You would have thought that this would normally have the effect of making the atomic radius smaller, because a greater attraction will pull those electrons closer to the nucleus. But the question suggests that the zinc has the bigger atom.

Which really is the bigger atom?

My first thought was that the atomic radii given by the questioner were wrong - because that would make the problem disappear. If you do a quick Google search, you will find a lot of variability for both zinc and copper - I quickly found values for radii ranging from 0.128 to 0.157 nm for copper and from 0.133 to 0.153 nm for zinc. I have no idea what the "correct" values are.

In some data lists, copper is the smaller atom; in others, zinc is smaller.

What if zinc is smaller than copper?

If zinc is the smaller atom, the problem would seem to disappear - you would have an atom with an ionisation energy greater than copper and an atom which is smaller. That would fit the usual explanations.

What if zinc is bigger than copper?

If zinc is bigger than copper, then at first sight there would seem to be a real difficulty here. The ionisation energy of zinc is bigger than copper's. That means that the outer electrons are being more firmly held. And yet some data shows that the zinc atom is bigger. That looks contradictory.

A possible solution

If zinc atoms are really bigger than copper atoms then I think I may have a possible way around the problem. This is, however, just speculation - I have no evidence for it.

When you measure or discuss ionisation energy you are thinking about removing electrons from isolated atoms in the gas state. However, metallic radius is found from the distance between atoms in a metal crystal. In other words, far from being isolated, the atoms are actually bound tightly to each other.

In the case of the transition metals (apart from zinc, which isn't properly classed as a transition metal at all), the metallic bonding involves some at least of the 3d orbitals as well as the 4s - that's why transition metals tend to have higher melting and boiling points than, say, Group 2 metals.

If some of the d electrons are involved in the bonding, then they can't also be involved in screening the outer (bonding) electrons from the nucleus - they are the outer bonding electrons. That means that for elements that use some of their d orbitals in their bonding, the measured atomic radius will be less than it would be in an unbonded atom. There would be more attraction from the relatively unscreened nucleus. If you are trying to compare trends in atomic radii with those in ionisation energies, you aren't working from the same essential electronic structures.

So, it seems to me that trying to relate the trend in ionisation energies to the trend in atomic (metallic) radii is actually pointless. It would only work if you had reliable van der Waals radii for the metal atoms - in other words, if they were in a non-bonded situation. (Follow this link if you aren't sure what a van der Waals radius is.)

The melting and boiling points of zinc are low for the rest of the series, reflecting the fact that it doesn't involve the 3d electrons in its bonding. That means that they should be fully available for screening purposes - even where the zinc is bonded. In the copper case, they are available for screening in an isolated atom, but some of them are not available when the copper is involved in a metallic bond. The effect of that is to pull the outer electrons closer to the nucleus when the copper atom is bonded - to give a smaller metallic radius, for example.

The smaller pull from the nucleus on the bonding electrons in zinc (because of the availability of all 10 d electrons for screening in the bonded atom) means that a bonded zinc atom would be bigger than a bonded copper atom. In other words, it may be possible to account for zinc having a bigger metallic radius than copper (if, in fact, it does!) using the same general principles that are used in explaining ionisation energies - but you need to dig around a lot.

I could, however, be completely wrong about this! If you have any reliable information about it (preferably with a reference) could you contact me via the address on the about this site page.

Like atomic radius and ionization energy, does ionic radius (for some particular charge, say 2+) follow the same pattern as atomic radius?

Here are the ionic radii for the 2+ ions that I have found from two different sources. All the values are in nm.

The sources:

Source 1: Chemistry Data Book 2nd edition; edited by Stark and Wallace; published by John Murray

Source 2: Nuffield Advanced Science Book of Data; published by Longman

There is little obvious similarity between the figures from the two sources, and neither shows any clear trend. What you can say (which is all the syllabus mentioned above is asking) is that the values don't change very much across the transition series.

What might you expect to happen to the size of the 2+ ions as you went across the series? Let's start by having a look at the electronic structures of all these ions. In each case, the original atom will have lost 2 electrons - and these always come from the 4s level before any 3d electrons get lost.

As you go across the series, there is an additional proton in the nucleus of each of the ions. All the outer electrons are in the same kind of orbitals, and there is no change in the amount of screening - in each case, the 3d electrons will be screened by the 1s, 2s, 2p, 3s and 3p electrons.

The net effect of this is that the attraction of the nucleus increases across the series and so you would expect the ionic radius to get smaller. But it doesn't - at least not all the way across the series.

So what is going wrong? I suspect that it is as simple as the fact that the ionic radius values being quoted aren't for isolated ions. They are going to be attached to something. They will either be surrounded directly by negative ions or will be covalently bound to ligands in a complex ion.

As soon as you put something else close to the positive ion, you will cause distortions in its electronic structure (particularly of the 3d orbitals) which means that the situation suddenly gets a lot more complicated - certainly beyond anything you will need for this level. This may well account for the differences between the ionic radius values from my two sources - they may be measured under subtly different conditions.

Once again, of course, I may be completely wrong about this! And again, if you have any reliable information about it (preferably with a reference) could you contact me via the address on the about this site page. But please, don't spend ages explaining the solution to me if it isn't capable of being understood by an intelligent 18 year old chemistry student, because I couldn't use it - even if I could understand it anyway!

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