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.
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$\begingroup$ Francium has a slightly greater ionization enthalpy than Cs. This could due to the Relativistic Effects $\endgroup$– user1811Jan 7 '14 at 14:25
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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. $\endgroup$– Nicolau Saker NetoJan 7 '14 at 14:34
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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.
answered Aug 15 '14 at 15:00