Video: Correlation between Ionization Energy and Electron Configuration of Aluminum

The 4th ionization energy of aluminum is approximately 154 eV. From which orbital is the electron being removed?

03:06

Video Transcript

The fourth ionization energy of aluminum is approximately 154 electron volts. From which orbital is the electron being removed?

The easiest way to answer this question is to take a look at the periodic table. When we look closely at our periodic table, we can see that aluminum is in period three and group 13. This question is asking us about electrons in orbitals around aluminum. So let’s use our periodic table to write out the electron configuration of aluminum. We can write out the electron configuration by counting along the periodic table.

In the first period, there are two elements. This puts two electrons in that first shell. This is an s shell. Next, it gets a little more complicated. In period two, we have two elements in the s block, the block on the left. This gives us two electrons in the s shell. Next, we have six elements in the p block, which gives us six electrons in the p shell. And it’s the 2p shell because we’re in period two.

Next, we have period three. Here, we have two elements in the s block of period three. So this gives us two electrons in the 3s shell. When we then look to the p block on the right-hand side of period three, we can see that aluminum is the first element. This means that, in the 3p shell, we only have one electron. So this is our complete electron configuration.

We’re specifically asked about which orbital the fourth ionization energy is from. To reach the fourth ionization energy, we must first go through three other ionization processes. Each time we remove an electron from aluminum, we produce an ionization energy. So let’s count backwards along the electron configuration.

The first ionization energy is the removal of the 3p1 electron. When we remove a second electron, this is from the 3s shell. The third time we remove an electron, it’s also from this 3s shell. And finally, we come to remove the fourth electron. And we can see it’s from the 2p shell. So this is our final answer.

The fourth ionization energy of aluminum corresponds to the electron being removed from the 2p orbital.

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