# Sorting ions by radius and ionization energy

Sort the ions $$\ce{F-}$$, $$\ce{Mg^2+}$$, $$\ce{O^2-}$$ by their ionic radius and the minimal energy needed to remove an electron.

For the radius I considered that they all have the same electron configuration – the one of $$\ce{Ne}$$ – but they have different amounts of protons. My though was, the more protons you have, the more all the outer electrons get "pulled" towards the core, reducing the radius.

For the energy I though it's basically the other way around, since the less everything holds together (aka the less the core charge), the easier it is to get another electron.

Radius: $$\ce{Mg^2+} < \ce{F-} < \ce{O^2-}$$
Energy: $$\ce{O^2-} < \ce{F-} < \ce{Mg^2+}$$

But apparently, the solution is:

Radius $$\ce{O^2-} < \ce{F-} < \ce{Mg^2+}$$
Energy $$\ce{F^-} < \ce{Mg^2+} < \ce{O^2-}$$

But I can't see why that should be.

• You thought right. – Ivan Neretin Jan 20 at 16:51