Z* effective charge and Ionization Energy

Question

I'm trying to figure out the patterns for Ionization Energies. I am familiar with the periodic trend, however things become quite different when we hit the 1st I.E. For example, Na has an I.E(1) of 495.8 kJ while its second I.E. rockets up to 4562 kJ while the atoms towards the right are much lower than this. The trend says that the I.E. increase up and right of the periodic table, which is not the case here.

My point is, in order to get a better estimate, would it be safe to say that the Zeff charge and atom size relate to the I.E?

Example: Effective Charges

Mg = 2 Al = 1 S = 4 Si = 2 Na = 1

We see that the strongest pull towards the charged center would be Al and Na in this case. However, having Al being the smaller atom would require more energy to remove the electron from its valence shell.

Now in the case of first Ionization Energy we have:

Mg = 3 Al = 2 S = 5 Si = 3 Na = 2

In this case, Na now has reduced its size due to the fact it jumped from n=3 to n=2 level and Al also reduced in size but is still a bigger atom than Na due to our trend.

My question is, is this approach fairly accurate or should I be looking somewhere else?

Answer 1

Using the common idiom: "full subshells are stable"

It is a little more compact to use the incorrect explanation and correct it than to explain in terms of correct. Also, when you hear this incorrect explanation, you will understand what is meant.

Yes, size is a factor (thus the "up" part- IE is higher in lower periods), but the observed anomaly in second-ionization energies in $\ce{Na}$ and $\ce{Al}$ is better related to removal of an electron from a full subshell.

$\ce{Na}$ has an electron configuration of $\ce{[Ne] 3s^1}.$ The first ionization gives $\ce{Na}$ makes it $\ce{[Ne]}$, the same as a noble gas- full 2p subshell. Because full subshells are stable, it takes a lot of energy to remove the first electron from it. $\ce{Al}$ is analogous; the second ionization means removing an electron from a full subshell.

The same is observed in the first ionization energies (in $kJ/mol$): $$\ce{Na}:495$$$$\ce{Mg}:737$$$$\ce{Al}:577$$ Ionization of $\ce{Mg}$ removes an electron from a full $\ce{3s}$ subshell, so it is a little higher than the trend.

There are many factors that contribute to exact values, but these are difficult to predict, which is why they taught a simple version in your class.

Correcting the myth

While it is common to refer to full subshells as being more stable, it isn't really what is going on. In reality, the next electron added in destabilized. This occurs because of shielding. Outer electrons feel repulsion from inner electrons, thus outer electrons are easier to remove than inner ones. The saying that "half-filled subshells are stable" is also used. Again, it is that the next added electron is destabilized, but in this case it is because the energy needed to spin pair the two electrons.

Answer 2

$Z_{eff}$ is certainly a major factor in determining ionization energies, however atomic and ionic radius probably shouldn't be viewed as having a direct causative relationship to ionization energy. It's more correct to say that ionization energy and atomic/ionic radius have some of the same underpinnings (namely, $Z_{eff}$ and various electron-electron interactions). Because many of the same underlying forces are at work, there is some correlation (i.e., ionization energy increases in the same direction as atomic radius decreases, certain specific anomalies aside), but don't confuse that correlation with causation.

As one moves down a group on the periodic table, $Z_{eff}$ obviously remains constant, however the number of core electrons shielding the nucleus increases, and consequently the valence level electrons become increasingly energetic as their distance from the nucleus grows. This fact largely accounts for the increase in atomic radius and the decrease in ionization energy that occurs as one moves down along any given group on the periodic table.

Conversely, as one moves right along a period, the number of core electrons remains constant, while the $Z_{eff}$ increases. It's reasonable to expect, therefore, that valence electrons will experience a stronger electrostatic attraction to the nucleus as one moves right along a period, causing both an increase in ionization energy and a decrease in atomic radius. The ionization energy trend mostly conforms to that expectation, with the notable exceptions of transitioning from group IIA to group IIIA, and from group VA to group VIA, where ionization energy (perhaps unexpectedly) drops. To explain these exceptions, compare the electron configurations:

• When removing the first group IIIA valence electron, it is being removed from a $p$ orbital, while the first group IIA valence electron would be removed from an $s$ orbital. Electrons in $p$ orbitals are somewhat more energetic due to the nuclear charge being partially shielded by the electrons of the preceding $s$ orbital (in addition to more complex quantum mechanical effects), hence they are easier to remove.
• The first group VIA electron to be removed is paired with another electron in the same $p$ orbital, while all $p$ orbital electrons of group VA elements are unpaired (in accordance with Hund's rule). Electron pairing causes some mutual electron-electron repulsion, making these electrons more energetic, resulting in a drop in ionization energy for group VI by comparison to group V.

As you move further down the periodic table, the contributions of electrons in $d$ and $f$ orbitals become significant, the energy gaps between subsequent principal energy levels get narrower, and the ionization energy trend becomes more strictly linear for main group elements (specifically, the exceptions I described above no longer apply once you reach principal energy level five).