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# Which element has a higher 3rd ionization energy, Al or Mg? Why?

Magnesium!

There is also an alternative way to approach this that is a bit different but builds on core principles you've been taught in General Chemistry, and distinguishes between the magnitude of increase in ionization energy:

- according to when one examines the ##Z_(eff)##
- according to when one goes against the .

As stefan mentions, knowing the electron configurations is a great idea for this question; it helps accompany how you visualize the electron distributions.

**THIRD IONIZATION ENERGY**

Just as the **first** ionization energy involves removing the **first** electron from the **highest** energy orbital, the **third** ionization energy involves removing a **third** electron from the **highest** energy orbital.

How might one express this? Well, if we let more "##Delta##" mean the application of heat and ##M## be a general metal:

##M(g) + Delta -> cancel(M^(+)(g)) + e^(-)## ##cancel(M^(+)(g)) + Delta -> cancel(M^(2+)(g)) + e^(-)## ##cancel(M^(2+)(g)) + Delta -> M^(3+)(g) + e^(-)## ##"---------------------------------"## ##M(g) + Delta -> M^(3+)(g) + 3e^(-)##

**The question is, which element is less likely to be dandy with losing an electron?**

**ELECTRON CONFIGURATIONS & OCTETS**

If you write out the electron configurations, note that ##Mg## and ##Al## are adjacent on the third period/row, and so the answer would have something to do with either the ##Z_(eff)## or a more substantial effect. Let's see...

##Mg##: ##1s^2 2s^2 2p^6 3s^2## ##Al##: ##1s^2 2s^2 2p^6 3s^2 3p^1##

or...

##Mg##: ##[Ne] 3s^2## ##Al##: ##[Ne] 3s^2 3p^1##

**Well, if we remove THREE electrons from each of these metals...**

...what we end up doing after removing the third electron is turning ##Al## into a ##Ne##-like cation. That's pretty favorable, because it gives the atom an **octet**, which you've been taught is in line with **stability** and **fairly inert qualities**. That's why the common oxidation state for ##Al## is ##Al^(3+)##.

However, with ##Mg##, what you end up doing is trying to remove the third electron from ##Mg^(2+)##, **which is already** ##\mathbf(Ne)## **-like**. That disrupts the stable octet it already has, and is unfavorable.

Hence, we would expect **a drastic increase in the third ionization energy of** ##\mathbf(Mg)## **compared to that of** ##\mathbf(Al)##, and there is!

##"IE"_(Mg)^((2)) = "1450.6 kJ/mol"## ##"IE"_(Mg)^((3)) = color(blue)("7732.6 kJ/mol")##

##"IE"_(Al)^((2)) = "1816.6 kJ/mol"## ##"IE"_(Al)^((3)) = color(blue)("2744.7 kJ/mol")##