Periodic Trend: Successive Ionization Energies - Video Tutorials & Practice Problems
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Successive Ionization Energies deals with the removal of multiple electrons from a gaseous atom or ion through multiple steps.
Successive Ionization Energies
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concept
Periodic Trend: Successive Ionization Energies
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Now the first ionization energy, abbreviated as I e sub 1, is the energy absorbed to remove the first electron from a gaseous atom. So remember, removing the electron makes nitrogen now positively charged. It's still in its gaseous phase, and here's the electron we removed. That has associated with this ionization energy of 1402, which will be in kilojoules. Now with more energy you can remove additional electrons in successive ionizations. So you can remove the second electron from n a plus to give to give us ionization energy 2. You could try to remove the third one, which would require ionization energy 3, and so on. Now successive ionizations is the removing of additional electrons in stages instead of all at once. So if I'm trying to remove 3 electrons from nitrogen I can't move them remove them all at the same time. I have to remove them 1 by 1. 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 it becomes 2+, and here's the electron I removed. So remember, as I remove each electron the ion becomes more positively charged. Then I need to remove the 3rd electron. I've already moved the first two, so at present I'm n+2. I'm about to remove the 3rd electron and when I do that I become 3+. And here goes the electron I just removed. So this would be ionization energy 1, ionization energy 2, and then ionization energy 3, where we remove the electron in 3 separate stages.
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example
Periodic Trend: Successive Ionization Energies Example 1
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Here we need to provide the 4th ionization energy equation for a manganese atom. So if we're doing the 4th ionization energy, we must assume we've already lost the first three electrons. So manganese would start in a 3 plus state. And remember, this happens with gaseous atoms, so it would be in its gaseous state. We're now gonna remove the 4th electron. When I remove the 4th electron it becomes 4+4's charge plus the electron, the 4th electron that I've just removed. This would represent the 4th ionization energy equation for the manganese atom.
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concept
Periodic Trend: Successive Ionization Energies
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Now with successive ionization energies, realize that there's 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 7th. And you can see that with each ionization energy for each element, the number increases. It gets harder and harder to remove the next electron. You can see Lithium goes from 520.2 to 7,298. Now we're gonna say traditionally, elements lose valence electrons to become isoelectronic to the noble gases. We're gonna say here a large jump in ionization energy results when we begin to remove the inner core electrons. Now think about it. Lithium is in group 1 a. It only wants to lose 1 electron to become just like a noble gas. When we try to take away its second electron, we're gonna need ionization energy too. Look at how much more energy is required. It's such a huge jump, more than a tenfold increase. Beryllium wants to lose 2 electrons to become like a noble gas because it's in group 2 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 it's now a noble gas. If I try to go in and take away the 3rd electron, look at how much the energy increases, it jumps way up. So you can predict where a 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 6 electrons it would 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 6th electron, it's going to cause a huge, it's gonna become a noble gas. But once we try to go in and take away that 7th electron, it's 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'd be and then we're a little bit greedy and try to go back in and take away one additional electron. This causes a spike in the ionization energy.
After an element gains an a noble gas electron configuration, there will be a large jump in ionization energy to remove the next electron.
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example
Periodic Trend: Successive Ionization Energies Example 2
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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 and c because they're in groups 1 a. They only wanna lose 1 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 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 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 3 a, and then beryllium and magnesium are on the other side of the periodic table in terms of group 2 a. 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.
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Problem
Problem
Which of the following represents the third ionization of Mn?
A
B
C
D
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Problem
Problem
Of the following atoms, which has the largest third ionization energy?