Periodic Trend: Successive Ionization Energies - Online Tutor, Practice Problems & Exam Prep

Ionization energy, denoted as $E_1$ for the first ionization energy, refers to the energy required to remove the first electron from a gaseous atom, resulting in a positively charged ion. Successive ionization energies involve removing additional electrons one at a time, with each subsequent ionization requiring more energy as the ion becomes more positively charged. This increase in energy is significant when inner core electrons are removed after the valence electrons have been stripped away, as seen in the large jumps in ionization energy values. These jumps can help predict the number of valence electrons in an element, as elements tend to lose electrons to achieve a noble gas electron configuration. For instance, lithium (in Group 1A) has a large increase in ionization energy after the first electron is removed, reflecting its desire to lose only one electron to resemble a noble gas. Similarly, oxygen would experience a significant increase after losing six electrons, corresponding to the number needed to reach a noble gas-like state.

Periodic Trend: Successive Ionization Energies

Now the first ionization energy, abbreviated as IE₁, is the energy absorbed to remove the first electron from a gaseous atom. So remember we're moving. 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 1402, which will be in kilojoules now.

With more energy you can remove additional electrons in successive ionizations. So you can remove the 2nd electron from Na⁺ 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.

If I'm trying to remove 3 electrons from nitrogen, I can't move 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 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 third electron. I've already moved the first two, so at present I'm N²⁺. I'm about to remove the third electron and when I do that I become 3⁺ and here goes the electron I just removed. So this would be ionization energy one, ionization energy two, and then ionization energy 3 where we remove the electron in three separate stages.

Periodic Trend: Successive Ionization Energies Example 1

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 1st 3 electrons. So manganese would start in a three plus state.

And remember this happens with gaseous atom so it would be in its gaseous state. We're now going to remove the 4th electron. When I remove the 4th electron, it becomes 4 plus fourths charge plus the electron, the 4th electron that I've just removed.

This would represent the 4th ionization energy equation for the manganese atom.

Mn ^ 3 + ( g ) → Mn ^ 4 + ( g ) + e -

Periodic Trend: Successive Ionization Energies

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 7298.

Now we're going to say traditionally elements lose valence electrons to become isoelectronic to the noble gases. We're going to 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 1A. 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 going to need ionization energy too. Look at how much more energy is required. It's such a huge jump, more than a 10-fold increase.

Beryllium wants to lose 2 electrons to become like a noble gas because it's in Group 2A. Here you can see that going from the first ionization energy to the next, there's an increase. It's less than double. Once we remove the 2nd 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 a jump in ionization energy will occur.

Remember each of these elements want to lose enough electrons to become just like a noble gas. Once you lose too many electrons, it's going to 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 jump. It's going to become a noble gas. But once we try to go in and take away that 7th electron, it's going to be a huge cost, right?

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'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.

Periodic Trend: Successive Ionization Energies Example 2

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.

Which of the following represents the third ionization of Mn?

• A. Mn⁻ (g) + e⁻ → 2 Mn²⁻ (g)
• B. Mn²⁺ (g) → Mn³⁺ (g) + e⁻
• C. Mn²⁻ (g) + e⁻ → 2 Mn³⁻ (g)
• D.
  Mn (g) → Mn³⁺ (g) + 3 e⁻

Of the following atoms, which has the largest third ionization energy? 

• A. Al
• B. Ca
• C. K
• D. Ba
• E. Cs

The successive ionization energies for an unknown element are:

IE₁ = 896 kJ/mol IE₂ = 1752 kJ/mol IE₃ = 14,807 kJ/mol IE₄ = 17,948 kJ/mol

To which family in the periodic table does the unknown element most likely belong?

• A. 1A
• B. 2A
• C. 3A
• D. 4A
• E. 5A

Here’s what students ask on this topic:

What is the first ionization energy?

The first ionization energy is the amount of energy required to remove the most loosely bound electron from a neutral atom in its gaseous state, to form a cation. It is usually expressed in electronvolts (eV) or kilojoules per mole (kJ/mol). The process can be represented by a chemical equation, where 'X' represents an element:

$X\left(g\right)\to X^{+}\left(g\right)+e-$

This energy is a measure of the force holding the electron within the atom and is indicative of how an element will react chemically. Elements with low first ionization energies tend to lose their outer electrons easily and are typically metals. In contrast, elements with high first ionization energies hold onto their electrons more tightly and are often nonmetals. The first ionization energy generally increases across a period (from left to right) in the periodic table and decreases down a group (from top to bottom).

What is the second ionization energy?

The second ionization energy refers to the amount of energy required to remove the second electron from an atom after the first electron has already been removed. This process results in the formation of a doubly charged ion from a singly charged ion. It is important to note that the second ionization energy is always higher than the first ionization energy. This increase in energy is due to the fact that, after the first electron is removed, the remaining electrons experience a greater effective nuclear charge—meaning the attraction between the nucleus and the electrons is stronger since there are fewer electrons to shield the positive charge of the nucleus. As a result, more energy is needed to overcome this attraction and remove an additional electron. Understanding ionization energies is crucial in the study of chemical reactivity and the formation of ions in various chemical processes.

Why do successive ionization energies increase?

Successive ionization energies increase because it becomes more difficult to remove an electron from an atom as you take more away. When you remove the first electron, the remaining electrons are held by the positive charge of the nucleus with a relatively weaker force since there are more electrons to spread the positive charge over. However, once you remove an electron, the positive charge of the nucleus is spread over fewer electrons, which increases the effective nuclear charge on each remaining electron. This means that each subsequent electron is held more tightly by the nucleus, and thus, more energy is required to remove it. Additionally, as you strip away electrons, the electron you are trying to remove may be closer to the nucleus and in a lower energy level, which further increases the ionization energy required to remove it.