Why is the second ionization energy of an alkali metal disproportionately high?

Jun 8, 2017

The ionization energy is expected to GREATLY INCREASE when we begin to remove NON-valence electrons..........

Explanation:

Sodium, $Z = 11$, has an electronic configuration of $1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{1}$. Should it undergo ionization........we can measure the energy associated with the transition........

$N a \left(g\right) + {\Delta}_{1} \rightarrow N {a}^{+} \left(g\right) + {e}^{-}$

The next electron, whose removal would give $N {a}^{2 +}$, derives from a non-valence shell, which is CLOSER to the nuclear core, and hence electrostatically more difficult to remove. In any case, on purely electrostatic grounds, for the second ionization.........

$N {a}^{+} \left(g\right) + {\Delta}_{2} \rightarrow N {a}^{2 +} \left(g\right) + {e}^{-}$

...${\Delta}_{2} \text{>>} {\Delta}_{1}$, $\text{a priori}$, because it should be harder to remove an electron from a positively charged ion than from a neutral atom.

For aluminium ion, $A {l}^{3 +}$, the three valence electrons have been removed. The fourth electron must be removed from a non-valence inner shell orbital, from the electronic core, and its removal is much more energetically difficult.

And please note that actual physical data, successive ionization energies, and electron affinities, underly the description of the electronic structure of atoms.