Periodicity III: ionization energy (2) - the Exceptions

First IE of elements in period 2

Let's take a look at the first IE of elements in period 2.

It is noticed that B shows a lower IE₁ as compared to Be, and IE₁ of O is also lower than that of N, despite the general trend of increasing IE₁ across the period as we have predicted in the previous post.

So is this some random phenomenon or does it have any scientific explanation? If it does have a reason behind, is it applicable to other periods as well? Let's take a look.

Orbitals

When an electron is being removed from an atom, the electron configuration of the atom would be altered. As a result, we would need to take a closer look at the orbitals and the electrons.

Be and B

Let's write down the electron configuration of beryllium and boron first.

Be: \( \ce{[He] 2s^2} \)

B: \( \ce{[He] 2s^{2} 2p^1} \)

So now we're going to remove one electron from these two atoms each. For Be, one electron in the 2s orbital would be removed, while the electron in 2p orbital of B is being removed.

We have learnt that in the same energy level, p sublevel has a higher energy than s sublevel (or p is farther away from the nucleus). Therefore, the electron to be removed in boron would required less energy to remove it. Just like, an apple on the second floor requires less energy to be carried to the third floor as compared to an apple that sits on the ground level, isn't it?

That explains why a lower IE₁ is observed for boron as compared to beryllium. Does it apply to period 3 elements then?

If we look at the elements in period 3 but in the same groups as beryllium and boron, Mg and Al, we would write down their electron configurations first.

Mg: \( \ce{[Ne] 3s^2} \)

Al: \( \ce{[Ne] 3s^{2} 3p^1} \)

So again we see the same scenario that electrons are being removed from 3s and 3p orbitals. We would thus expect the electron in Al to require less energy to be removed. Is it true?

Scientists tell us that IE₁ of Mg is 738 kJ/mol while that of Al is only 578 kJ/mol.

Since our hypothesis is able to successfully predict the IE₁ of other elements, we would be confident that our hypothesis is correct.

N and O

It is slightly different for nitrogen and oxygen because both atoms would have the electrons in their 2p orbitals to be removed.

N: \( \ce{[He] 2s^2 2p^3} \)

O: \( \ce{[He] 2s^{2} 2p^4} \)

We would turn to the box-and-arrow diagram to have a clearer view of the 2p orbitals, and the electron to be removed is highlighted in red.

N: \[ \displaylines{ \boxed{ \text{ } \upharpoonleft \text{ } \mid \text{ } \upharpoonleft \text{ } \mid \text{ } \textcolor{red}{\upharpoonleft} \text{ } } \newline \text{2p} } \]

O: \[ \displaylines{ \boxed{ \text{ } \upharpoonleft \textcolor{red}{\downharpoonright} \text{ } \mid \text{ } \upharpoonleft \text{ } \mid \text{ } \upharpoonleft \text{ } } \newline \text{2p} } \]

As you might have already noticed, removing the electron from the nitrogen atom disrupts the stable half-filled p orbital, while removing the electron from the oxygen atom would give rise to the stable half-filled p orbital configuration.

Therefore, nitrogen would require more energy to remove that electron in order to break the stability. On the other hand, oxygen requires less energy to remove that electron as it already has the tendency to lose that electron, just like an apple on the table has the tendency to drop onto the floor so that we don't need to input any effort to do so.

Another reason that oxygen has a lower IE₁ is that there are initially two electrons in the orbital whereby the repulsion between the electrons would do us a favor to push the red electron out. That saves us energy too.

These two reasons, the stability of the half-filled p orbital in nitrogen, as well as the repulsion between the electrons in the same orbital in oxygen, lead to an unusual decreasing of IE₁ across the period.

Stable electron configuration

Apart from half-filled orbitals, fully filled orbitals are also more stable as compared to other electron configurations.

This is because of the symmetry of electrons in both the half-filled and fully filled electron configurations. It gives rise to some balance in the interactions between electrons and between electrons and nucleus.

With this knowledge, could you explain why in period 3, phosphorous has a lower IE₁ than sulfur?

Summary

Let's wrap up everything.

Firstly, IE₁ of atoms would decrease down the group, due to the fact that a greater number of energy levels results in a longer distance between the nucleus and valence electrons, and hence a weaker attraction force.

Secondly, IE₁ of atoms across the period would generally increase. These atoms have the same number of energy levels, therefore the distances between the nucleus and valence electrons are more or less the same (in fact the distances decrease across the period). As more protons exist in the nucleus, a greater attraction between the nucleus and the electron could be expected. Thus, more energy would be needed to remove electrons from the atoms with a larger atomic number.

However, Be and Mg have higher IE₁ as compared to B and Al respectively, due to the fact that they will have their s electrons removed but the latter ones would have the p electrons being removed instead.

N and P also have higher IE₁ than O and S do. This is related to the stability of half-filled p orbitals in N and P atoms, and the repulsion between electrons in the same orbital for O and S atoms.

One last thing that we'd like to highlight is that all noble gases have very high IEs because they are pretty stable and it's difficult to remove electrons from noble gas atoms.