[mac227]

Why is there a significant increase in the twelfth ionization energy for aluminum?

I know ionization energy levels increase as you get into the higher electron shells, but I do not understand this question.

Can comeone help me understand this concept because I also need to find where similar increases in ionization energies occur in Mg and Si.

[DevaDevil]

do you know the electron configuration of aluminium?

[mac227]

1s2 2s2 2p6 3s2 3p1

SO there is a large increase in ionization energy because the 3s2 orbital is full?  At this point there would be 12 electrons total so there it takes a lot of ionization energy to add an electron to the 3p1 orbial??

[DevaDevil]

Indeed: 1s2 2s2 2p6 3s2 3p1
the first (13th) electron to go is the lone 3p electron.

After this: Al⁺: 1s2 2s2 2p6 3s2

The 12th is a 3s electron, not just one electron lower, but a full electron shell lower. This means it is a lot tighter attracted to the core and thus the huge increase in ionization energy.

[mac227]

Ok, I think i understand now.

So as the orbital shells increase (to 3s and 3p) the electrons are getting closer to core--making it harder and harder to keep adding electrons.  SO for example aluminum is
1s2 2s2 2p6 3s2 3p1

So it will take even more ionization energy to get from the 3s2 to 3p1 orbital? and so on

And it would even take more energy to add an additional electron to fill the 3p orbital?  As is the case for silicon--
1s2 2s2 2p6 3s2 3p2

[DevaDevil]

it keeps getting harder and harder to remove electrons as you go to shells closer to the core.

In silicon (1s2 2s2 2p6 3s2 3p2), removal of the 1st and 2nd electrons is similar in energy, then a large increase in energy to remove the first 3s electron.