Z* effective charge and Ionization Energy 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?
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.