# Why does radium have a higher first ionisation energy than barium?

I'm wondering why radium appears to buck the general trend that first ionisation energies decrease as you move down a group in the periodic table: barium (the group 2 element preceding it) has a first ionisation energy of $$\pu{502.9 kJ/mol}$$, whereas radium has a slightly higher first I.E. of $$\pu{509.3 kJ/mol}$$ (from the Wikipedia, although my textbook agrees).

Is there any explanation for this at present? (I imagine that quantum mechanics may be involved in some way, but I'm not entirely sure how)

There are most likely other examples of the trend being broken, but this is the only one I've come across so far and I'm curious to know why this is the case.

• The same trend-breaking can be observed with Cs and Fr. Also the trend of decreasing first ionisation energy doesn't seem to hold with the transition from 5th period to 6th period transition metals (as well as some of the main group elements) and the transition from lanthanides to actinides (see: chemreference.com), so I guess it must have something to do with the additional f electrons, but I have no ideas what it is exactly. – Philipp Dec 20 '12 at 14:05

I think it's also important to mention relativistic effects here. They already start becoming quite visible after $$Z=70$$, and $$\ce{Ra}$$ lies a good bit after that.

In very heavy atoms, the electrons of the $$\ce{1s}$$ orbital (actually, all orbitals with some electron density close to the nucleus, but the $$\ce{1s}$$ orbital happens to be the closest and therefore most affected) are subjected to very high effective nuclear charges, compressing the orbitals into a very small region of space. This in turn forces the innermost electrons' momenta to be very high, via the uncertainty principle (or in a classical picture, the electrons need to orbit the nucleus very quickly in order to avoid falling in). The momenta are so high, in fact, that special relativity corrections become appreciable, so that the actual, relativistically corrected momenta, ($$p_{\text{relativistic}}=\gamma p_{\text{classical}}$$) are somewhat higher than the approximate classical momenta. Again via the uncertainty principle, this causes a relativistic contraction of the $$\ce{1s}$$ orbital (and other orbitals with electron density close to the nucleus, especially $$\ce{ns}$$ and $$\ce{np}$$ orbitals).

The relativistic contraction of the innermost orbitals creates a cascade of electron shielding changes among the rest of the orbitals. The final result is that all $$\ce{ns}$$ orbitals are contracted, getting closer to the nucleus and becoming shifted down in energy. This is relevant to the question because the $$\ce{7s}$$ valence electrons in $$\ce{Ra}$$ are more attracted to the nucleus than one would expect from a simple trend analysis, since they rarely take into account the increase of relativistic effects as one goes down the periodic table.

Thus, the first (and second) ionization energy of $$\ce{Ra}$$ becomes higher than expected, to the point that there's actually a upward blip in the downward trend. Eka-radium ($$Z=120$$) would have far stronger relativistic effects, and can be expected to have a significantly higher ionization energy compared to $$\ce{Ra}$$. In fact, relativistic effects will conspire to make the group 2 metals slightly more noble! Though the periodic table becomes such a mess near the super heavy elements that it's hard to say whether it'll be a clearly visible trend, or just one effect to be combined with several others.

• This is a good theoretical explanation of the effects noted in my answer. Basically, any element after the lathanides will have a higher first ionization energy than the one above it for this reason. – Ben Norris Apr 28 '13 at 23:10
• Well, this is actually not the case for heavy elements in groups 15-18 (Bi, Po, At, Rn and those below them). Here, relativistic effects actually destabilize two of the three valence $np$ orbitals (technically, the $np$ subshell has its degeneracy broken due to strong spin-orbit coupling, forming a stabilized $np_{1/2}$ orbital and two destabilized $np_{3/2}$ orbitals), so that even with increased effective nuclear charge, all ionization energies up to $np_{1/2}^2$ require less energy than for the elements above. This effect is most apparent in Bi, because Pb has slight relativistic nobility. – Nicolau Saker Neto Apr 29 '13 at 1:34
• "relativistic contraction" I am not sure what you mean by contraction?. Is it like the probability of finding electron shrinks to a smaller space? Or the orbitals become closer to the nucleus? – Mockingbird Jul 27 '17 at 8:20
• @Mockingbird I'm not too sure what the difference is between the two options you've given, but basically the average distance of the electron to the nucleus, in all $s$ orbitals, is smaller when you include relativity. – Nicolau Saker Neto Jul 27 '17 at 8:41

This behavior can be attributed to the same phenomenon as that which causes the lanthanide contraction. The electrons in $$f$$ subshells are very poor at nuclear shielding, so the $$s$$ electrons in the next higher shell are closer (on average) to the nucleus than you might expect. If these electrons are closer to the nucleus, then the atom exhibits a smaller radius than expected and these $$s$$ electrons are more difficult to remove. According to the Wikipedia article on the lanthanide contraction, there are both quantum mechanical and relativistic causes for the lanthanide contraction.

This behavior also occurs in hafnium (72), which is right after the lanthanides. Hafnium's first ionization energy is $$\pu{658.5 kJ/mol}$$, while zirconium (40, right above hafnium) is $$\pu{640.1 kJ/mol}$$. Barium is before the lanthanides and radium is after them. Any pair so split will probably exhibit this new trend in ionization potentials. Go and check it out!

Elements 57+ have electrons in f orbitals.

Radium has a full 4f sublevel these electrons as less effective at shielding than d electrons (f

• You can find more detailed information on the poor shielding of f block electrons here. chemwiki.ucdavis.edu/Inorganic_Chemistry/Descriptive_Chemistry/… – Julia Gonzalez Jan 28 '13 at 5:20
• Looks like your answer has been cut short. In addition, it would be nice to incorporate some of the link into your answer (writing in your own words would be nice, though quoting is fine) – ManishEarth Jan 28 '13 at 17:54

It is because the 4f orbital of radium shields the 7S orbital less effectively from nuclear attraction, therefore a higher ionization energy is required for their removal.

Well it's all about the shielding effect,f subshell have many holes and thus the nuclear influence on the outer s electrons increases,so valence electrons are much tightly bounded to the nucleas.hope you got the answer