Periodic Trend: Successive Ionization Energies

now the first ionization energy abbreviated as e sub one is the energy absorbed to remove the first electron from a gaseous Adam. So remember, removing the electron makes nitrogen now positively charged. It's still in its gaseous phase. And here is the electron we removed that has associated with this ionization energy of 14 02 which will be in killing jewels. Now with more energy, you can remove additional electrons in successive ionization so you can remove the second electron from an A plus to give to give us ionization energy to you could try to remove the third one, which would require ionization energy three and so on. Now successive organizations is the removing of additional electrons in stages instead of all at once. So I'm trying to remove three electrons from nitrogen. I can't move them. Remove them all at the same time. I have to remove them one by one. So here we have the first ionization energy of nitrogen. So if I was trying to remove the next electron, I've already removed the first. So now I'm at this plus one stage. I'm removing the second one so now becomes two plus. And here's the electron I removed. Remember, As I remove each electron, the ion becomes more positively charged. Then I need to remove the third electron. I've already moved the first two. So at present, N plus two, I'm about to remove the third electron. And when I do that, I become three plus. And here goes the electron I just removed. So this would be ionization energy, one ionization energy to and an ionization energy three, where we remove the electron in three separate stages.

here. We need to provide the fourth ionization energy equation for, um, agonies Adam. So if we're doing the fourth ionization energy, we must assume we've already lost the first three electrons, so manganese would start in a three plus state. And remember, this happens with gashes. Adams so would be in its gaseous state. We're now going to remove the fourth electron. When I remove the fourth Electron two becomes four plus four. It's charge, plus the electron, the fourth electron that I've just removed. This would represent the fourth ionization energy equation for the mag unease, Adam.

now with successive ionization energies, realized that there is an ever increasing amount of energy required each time an electron is removed. So if you look at the numbers here, you can see we have our first ionization energies, then our second all the way to our seventh. And you can see that with each organization energy for each element the number increases. It gets harder and harder to remove the next electron. You can see lithium goals from 5. 20.2 to 7298. Now we're gonna say traditionally, elements lose valence electrons to become isil. Elektronik to the noble gasses were going to say here, Ah, large jump in organization energy results when we begin to remove the inner core electrons now think about it. Lithium is in Group one A. It only wants to lose one electron to become just like a noble gas. When we try to take away its second electron, we're gonna need ionization energy to look at how much more energy is required. It's such a huge jump, more than a 10 fold increase. Beryllium wants to lose two electrons to become like a noble gas because it's a group to a here. You can see that going from the first ionization energy to the next, there's an increase. It's less than double when. Once we remove the second electron, it's now a noble gas. If I try to go in and take away the third electron, look at how much the energy increases. It jumps way up so you can predict where jump in ionization energy will occur. Remember each of these elements wanna lose enough electrons to become just like a noble gas. Once you lose too many electrons, it's gonna jump up in energy oxygen, for example, if it were to lose six electrons, it will become like a noble gas. I know oxygen likes to gain electrons, not lose them. But we're talking about ionization energy here. When it loses that sixth electron, it's going toe cause a huge um, it's gonna become a noble gas. But once we try to go in and take away that seventh Electron is gonna be a huge cost. Alright. So again we can predict where this big jump in ionization energy will be. It happens once our element has reached noble gas status, and then we'll be there were a little bit greedy and try to go back in and take away one additional electron. This causes a spike in the ionization energy.

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.