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### Course: AP®︎/College Chemistry>Unit 1

Lesson 6: Periodic trends

# Worked example: Identifying an element from successive ionization energies

When electrons are removed in succession from an element, the transition from removing valence electrons to removing core electrons results in a large jump in ionization energy. By looking for this large jump in energy, we can determine how many valence electrons an element has, which in turn can help us identify the element. Created by Sal Khan.

## Want to join the conversation?

• I get the idea of needing more energy to remove the core electrons. But once you remove the electron from the 2nd shell, wouldn't the next electron be easier to remove. For that matter, shouldn't the first 3 removals take less and less energy, and then have a big spike to remove the 4th electron, and then less to remove the fifth?
• No, because electrons on the same shell also repel each other, so the more you have, the easier it is to remove one.
• I haven't received and answer yet for my question 3 days ago, and this is the revised version:

Why can't the identity of the element be P? Though s and p orbitals are in the same shell, the p orbital is supposed to have higher energy. Please check my reasoning below:

1. When you add 3p1, the outermost electron is in the 3p orbital (even though 3s and 3p are all valence electrons).
2. Therefore the 3p1 electron is subject to electron-electron repulsion by 3s2 electrons
3. The distance between the nucleus and the electron in 3p is further than the 3s2 electrons
By using 1,2,3 in Coulombs's law, when you add 3p1, ionization energy decreases. Which also makes sense because a lone electron farthest away from the nucleus should be easiest to pull away.
In reverse therefore, when you take away 3p1, ionization energy increases, and that could account for the leap between the third and fourth IE.

• When we look at the changes between different ionization energies, we see an abnormally large increase from the third to the fourth ionization energies. Much higher than the transition from any of the other ionization energies. Since ionization energy is the amount of energy required to remove an electron, a high ionization indicates a difficult to remove electron. This tells us the atom resisted losing that fourth electron more so than the first three. The reason for this resistance is because the atom reaches a relatively stable electron configuration by losing the first three electrons and does not want to disturb that stability by losing a fourth electron.

Such a large difference between the third and fourth ionization energy corresponds to the stability associated with a noble gas electron configuration. So we’re looking for a third period element which has a cation of +3 charge with the same electron configuration as the previous noble gas, neon. This can only be aluminum then.

There is an energy difference between the 3s and 3p subshells, but it’s slight and this wouldn’t correspond to such a large energy difference as the one between the third and fourth ionization energies here. The main culprit for huge energy difference between ionization energies is the cation achieving a noble gas electron configuration and not wanting to lose that stability.

For phosphorus we would expect a large difference between the fifth and sixth ionization energies, not between the third and fourth. To confirm this here are the first seven ionization energies of phosphorus in kJ/mol: 1011.8, 1907, 2914.1, 4963.6, 6273.9, 21267, 25431; and we can see that abnormally large difference between the fifth and sixth ionization energies as expected. And like aluminum, its +5 cation has the same electron configuration as neon and wants to remain as such.

Side note, you don’t need to post the same question multiple times to get a response. People will respond when they have time, not because you repeat it. Doing so makes you look needy.

Hope that helps.
• If the period of the element isn't given, how would we determine what element it is?
• If you narrow down the options to a group 13 element then you would need to know the ionization energies of that group’s elements, or their relative amounts. Ionization energy within a group generally decreases as you move from top to bottom. This is because as you move down the valence electrons are located in shells farther away from the nucleus and so feel less attraction to the nucleus and are easier to remove. This means elements like boron and aluminum will have similar ionization energy patterns which show them to be in the same group, but boron will have a higher first ionization energy than aluminum.

So without actually providing the ionization energies for all the group 13 elements, they could say that the element has the second highest first ionization energy in its group, which would be aluminum. But even that wouldn’t work well since gallium (the element beneath aluminum) has about the same first ionization energy as aluminum. The general rule is that ionization energy decreases down a group, but the pattern isn’t as identifiable as the pattern across a period.

Most questions wouldn’t bother with all this nuance and would simply just say which period the element is in.

Hope that helps.
• What about a third period element with the numbers:
IE1 1230 kJ/mol
IE2 2400 kJ/mol
IE3 3600 kJ/mol
IE4 8000 kJ/mol
IE5 10500 kJ/mol
IE6 22000 kJ/mol
IE7 27500 kJ/mol
IE8 35200 kJ/mol
(1 vote)
• The biggest jump (change) is between IE5 and IE6 so I would guess that from IE6 & up we are removing core electrons.

This means that we most likely have 5 valence electrons, meaning we would be in the 15th group on the periodic table, meaning we would guess Phosphorous (P).

I'm not sure if you were challenging me or asking a question but I hope this helps & is beneficial to someone. :-)
• Why can't the identity of the element be P? Though s and p orbitals are in the same shell, the p orbital is supposed to have higher energy. Please check my reasoning below:

1. When you add 3p1, the most outer electron is in the 3p orbital (even though 3s and 3p are all valence electrons).
2. Therefore the 3p1 electron can be shielded by 3s2 electrons
3. The distance between the nucleus and the electron in 3p is further than the 3s2 electrons
By 1,2,3, when you add 3p1, ionization energy decreases. Which also makes sense because a lone electron farthest away from the nucleus should be easiest to pull away.
In reverse therefore, when you take away 3p1, ionization energy increases, and that could account for the leap between the third and fourth IE.

• Although the 3p orbital has a higher energy level than the 3s orbital, the 3s orbital does not shield the 3p. The 3p orbital has higher energy because of the warped nature of the p-orbital, not because the 3p orbital is any further from the nucleus than the 3s orbitals. Additionally, adding 3p1 would not decrease the ionization energy. When you add valence electrons, ionization energy generally increases because of a decreased shielding effect.
• Why can't Boron or Gallium be an acceptable answer as well?
• "The first five ionization energies for a third-period element are shown below", key word being third-period element. Boron and Gallium have the same trend, but are in the second and fourth periods respectively. Additionally the numbers given are only true for aluminium. Boron and Gallium have different values.

Hope that helps.
• Why is there a difference between the first, second, and third ionization energies? For the first three electrons removed, they were all valence electrons so shouldn't they all have been equally as easy to remove?
(1 vote)
• Subsequent ionization energies of all elements have a general increasing trend because it becomes progressively harder to remove additional electrons. When we remove the first electron from a neutral atom with the first ionization energy, it becomes a cation, or a positively charged ion. Having an overall positive charge means the remaining electrons feel a greater force of attraction to the atom. Essentially by removing one electron, the remaining electrons have one less repulsive force (because electrons repel each other with their negative charge) but the same attractive force (from the positive protons) resulting in a net increase of attractive force. Removing even more electrons will result in even higher third, fourth, fifth, etc. ionization energies because the magnitude of the positive charge continues to increase.

Hope that helps.