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

Solution 1:

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.

Solution 2:

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!

Solution 3:

Elements 57+ have electrons in f orbitals.

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