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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.

Gaurang Tandon
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ankit
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1 Answers1

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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.

Jannis Andreska
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