# Can ionic and atomic radii be compared?

Can ionic and atomic radii be compared? For example, this is a common problem in some intro chem courses:

Rank the following from smallest to largest: sodium ion, magnesium ion, neon, fluoride, oxide.

Any student who has been trained in trends should be able to get the "answer", but if one looks at the actual figures, there are data that show that sodium ion is larger than neon atom.

I am of the opinion that it is comparing apples to oranges and that the problem can be rescued by removing neon from the list.

• I don't see how $\ce{Na^+}$ could possibly be larger than $\ce{Ne}$, at any level of theory. I think it's more of a case of atomic sizes being hard to measure (in fact there are several ways to define atomic/ionic radii, and measurements fluctuate within each definition). Are you sure you aren't comparing values from different definitions? – Nicolau Saker Neto Aug 23 '14 at 2:28
• My thought is that they cannot be compared. Is there a definition of radius that works for both ions and atoms? – Brinn Belyea Aug 23 '14 at 3:28
• See Ionic radius vs Atomic radius might help – Freddy Aug 23 '14 at 9:05

• We expect POSITIVE ions to be SMALLER than the corresponding atom because there are fewer electrons than protons => an excess of POSITIVE charge => the electron cloud is more attracted to the nucleus on average and becomes more 'concentrated' => $\ce{Na+}$ is 'small' and $\ce{Mg^{2+}}$ is 'very small'.
• We expect NEGATIVE ions to be LARGER than the corresponding atom because there are more electrons than protons => an excess of NEGATIVE charge => the electron cloud becomes more diffuse to compensate, lowering the Coulombic interactions => $\ce{F-}$ is 'large', $\ce{O^{2-}}$ is 'very large'.
Even though $\ce{Na+}$ has a lower principal quantum number for its outer shell than $\ce{Mg^{2+}}$, Mg is twice as charged and I believe this can be more important. $\ce{F-}$ and $\ce{O^{2-}}$ have the same principal quantum number for their outer shells, but $\ce{O^{2-}}$ is more charged and accordingly, we expect that $\ce{O^{2-}}$ has a larger radius.
$\ce{Ne}$ carries no unbalanced charge and has comparable valence principal quantum number to the ions, so we expect its radius to fall somewhere between the positive ions (which are smaller) and the negative ions (which are larger). Go over these arguments in your head and try to put the pieces together yourself from here.