Why is the ionization enthalpy of francium greater than that of cesium, even though it has a larger size? I found no Google result regarding this.

  • $\begingroup$ Francium has a slightly greater ionization enthalpy than Cs. This could due to the Relativistic Effects $\endgroup$
    – user1811
    Commented Jan 7, 2014 at 14:25
  • 3
    $\begingroup$ Here's a very similar question. The answer is the same, as barium and radium are also in the s-block, merely one element to the right of your pair. $\endgroup$ Commented Jan 7, 2014 at 14:34

1 Answer 1


The reason is the same as in the case of barium and radium, a contraction of core electron orbitals with increasing effective nuclear charge. This also changes the shielding (from nuclear charge) of electrons in outer orbitals, and in the end, all $s$ orbitals are contracted and lowered in energy. $s$ orbitals are particularly affected because they have no nodal plane at the nucleus.

This means that the single valence electron in the $7s$ orbital of $\ce{Fr}$ is closer to the nucleus and thus more tightly bound than the $6s$ electron of $\ce{Cs}$. More energy is therefore required to remove the $7s$ electron to form $\ce{Fr+}$ than in the case of $\ce{Cs}$, whose $6s$ electron is more loosely bound.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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