# Why is Aluminum's 2nd ionization energy higher than Silicon's?

According to the periodic table trend for I.E. Silicon should have a higher 2nd ionization energy than Aluminum so I'm confused.

Let us look at the electronic configurations of aluminum and silicon and their respective cations.

$\ce{Al}: [\ce{Ne}]\mathrm{3s^2\, 3p^1}$

$\ce{Si}: [\ce{Ne}]\mathrm{3s^2\, 3p^2}$

$\ce{Al^2+}: [\ce{Ne}]\mathrm{3s^1}$ It involves removing from $\mathrm{3s}$ sub shell

$\ce{Si^2+}: [\ce{Ne}]\mathrm{3s^2}$ It involves removing from $\mathrm{3p}$ sub shell

A $\mathrm{3p^1}$ electron has lower binding energy than that of $\mathrm{3s^2}$ electron.

It is due to more screening effect (as the predecessor sub-shells are higher in $\ce{Si}$) and higher stability attained by having a completely filled $\mathrm{3s}$ sub-shell

• Hi and welcome to chemistry.stackexchange.com. Feel free to take a tour of the site. I edited your post to include the \ce{...} and the \mathrm{...} syntaxes for MathJax. Learn more about them in the help center or this collection of editing hints. Nice first answer. – Jan Jan 25 '16 at 13:52