Which element has the highest second ionization energy: Li, K, or Be?

Video transcript

hey everyone in this example, we need to select the element with the largest third ionization energy. We want to recall that the electron configuration of an atom that is stable will correspond to stable core electrons and therefore correspond to a low value for the ionization energy. So beginning with Neon, we would recognize that Neon is in group eight a. Corresponding to eight valence electrons. And so we would recall that also Neon has the atomic number 10. So we should recall that atomic number is represented by the symbol Z. So we would say z equals 10. And we would also recall that For neutral atoms, our atomic number is equal to our number of electrons. So we would say that this is also equal to 10 electrons. So to calculate our core electrons for neon, we would take 10 minus R eight valence electrons which would equal a difference of two for our two core electrons. And so if we were to find our third ionization energy of the atom neon By removing three electrons from our configuration of neon, this would not affect our two core electrons. And so because our core electrons are unaffected, we would therefore have a low third ionization energy for neon. So continuing on with choice B. Because we would rule out choice A. Since we want to select the atom with a high third ionization energy, we're going to recognize that sodium is located in Group one A of our periodic table and this would correspond to one valence electron. We would also recognize that sodium has the atomic number or z value equal to 11 Corresponding to 11 electrons total. So we would take 11 -1 of our valence electrons to give us a total of 10 core electrons, meaning that if we were to calculate the third ionization energy of sodium by removing three of our 11 total electrons, we would affect two of our core electrons. So we have two core electrons affected creating instability in our configuration. And so we would have a high third ionization energy for sodium. So we won't rule out choice B yet we're going to move on to calculate for choice. See so for choice C. Magnesium We would recognize that magnesium is located in group two a of our periodic table which would therefore correspond to two valence electrons. We would recognize that Magnesium has the atomic number equal to a value of 12 Corresponding to 12 electrons. So we would have 12 electrons minus our two valence leaving us with 10 core electrons. So if we were to calculate the third ionization energy of magnesium, we would also have two of our core electrons affected, giving us therefore a high third ionization energy for magnesium. So we won't rule out choices yet. So now we would move on to calculate for choice d where we recognize that aluminum is located in group five a of our periodic table corresponding to five valence electrons. We would recognize that aluminum has a atomic number or z value equal to 13. So we would have 13 total electrons minus our five valence electrons giving us a difference of eight core electrons. And so if we were to calculate the third ionization energy for our atom aluminum, we would go ahead and remove three electrons from our configuration of aluminum, Meaning we would have 13 -3. And that would give us zero of our core electrons unaffected. And so therefore we would have a low third ionization energy for aluminum. So we can rule out choice D. We've already ruled out choice A and now we're just between B and C. So to decide between the ionization energy between magnesium and sodium, we want to recall that as an ion for sodium because it's in Group one A. We would form the N. A. Plus catalon. And because magnesium is in group two A of our periodic table, we would form the MG two plus catalon. Now, as you can see, we have a higher caddy in charge for magnesium. And so because we formed this MG two plus Catalan, we would say that therefore we have a higher electron attraction for magnesium. And so it's going to therefore have a higher ionization energy because it's going to be more difficult to remove this third electron for the third ionization energy of magnesium. And so our final answer is going to be choice C magnesium as the element with the largest third ionization energy. I hope that everything I explained was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.