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$"Aluminum, Z = 13"$, with an electronic configuration of $1s^(2)2s^(2)2p^(6)3s^(2)3p^1$, versus $"Natrium, Z = 11":$ $1s^(2)2s^(2)2p^(6)3s^(1)$ .

The second electron removed from the sodium atom is removed from an inner non-valence shell, which is closer to the nucleus, and thus more tightly held. It makes sense that this electron should require more energy to remove. On the other hand, aluminum has three , with $n=3$. The fourth ionization of aluminum should be disproportionately high.

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