Periodic Trend: Successive Ionization Energies Example 2

by Jules Bruno
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So here we're keeping the success of ionization energies chart up just for perspective. Here it says off the following atoms, which has the smallest increase for its second ionization energy. We should automatically eliminate options B and C because they're in groups one A. They only want to lose one electron to become a noble gas. And once they obtain that noble gas status going in to take away the second electron causes a big increase in ionization energy. We see that happening with with him here, trying to take away that second electron causes a huge spike in the cost. Now we're left with aluminum, magnesium and beryllium. Now taking away that second electron does not disrupt their noble gas status because they're not that yet. So then we just go by what we know in terms of ionization energy, as we had towards the top right corner. Ionization energy increases here. Aluminum is around, group is on group three A. And then beryllium and magnesium are on the other side of the periodic table in terms of group two way. So we know that based on that that aluminum, would it be our answer magnesium Since its most to the left. It would start off with the lowest ionization energy. And so we could assume here that we wouldn't see as big of a jump in its second ionization energy making option. D our best answer.