So here we're keeping the successive ionization energies chart up. Just for perspective. Here it says of the following atoms, which has the smallest increase for its second ionization energy, we should automatically eliminate options B&C because they're in groups 1A they only want to lose 1 electron to become a noble gas, and once they obtain that noble gas status, going in to take away the 2nd electron causes a big increase in ionization energy.

We see that happening with lithium 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 head towards the top right corner. Ionization energy increases here.

Aluminum is around group is on Group 3A 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 wouldn't be our answer. Magnesium. Since it's most to the left, it would start off with the lowest ionization energy, and so we can assume here that we wouldn't see as big of a jump in its second ionization energy, making option D our best answer.