Worked example: Identifying an element from successive ionization energies | Khan Academy

We are told that the first five ionization energies for a third period element are shown below. What is the identity of the element? So pause this video and see if you can figure it out on your own, and it'll probably be handy to have a periodic table of elements.

So before I even look at a periodic table of elements, let's make sure we understand what this table is telling us. This is telling us that if we start with a neutral atom of this mystery element, it would take 578 kilojoules per mole to remove that first electron to turn that atom into an ion with a plus one positive charge.

Then it would take another 1817 kilojoules per mole to remove a second electron, so to make that ion even more positive. After that, it would take another 2745 kilojoules per mole to remove the third electron. Then, to remove the fourth electron, it takes a way larger amount of energy. It takes 11,000 kilojoules per mole, and then the fifth electron takes even more: 14,842 kilojoules per mole.

For the first, second, and third, you do have an increase in ionization energy, but when you go to the fourth, the energy required to remove those is way higher. So to me, these look like you're removing valence electrons, and these look like you're removing core electrons.

One way to think about it is let's look at our periodic table of elements and look for a third period element that has three valence electrons. So we have our periodic table of elements. We want a third period element, so it's going to be in this third row. Which of these has three valence electrons?

Well, sodium has one valence electron, magnesium has two valence electrons, aluminum has three valence electrons. So one way to think about it is that first electron: it's a reasonable ionization energy. Then the second one, a little higher; then the third, a little bit higher than after that. But then the fourth, you're starting to go into the core. You're going to have to take an electron out of that full second energy shell, which takes a lot of energy.

So this is pretty clearly aluminum that is being described.