How does the ionic radius of a nonmetal compare radius?

1 Answer
May 16, 2017

Answer:

The ionic radius of a non-metal SHOULD be GREATER than the atomic radius........Why?

Explanation:

Metals are in general electron-rich materials, and their chemistry TENDS to be reducing. Because, in general, they come from the left hand side of the Periodic Table, nuclear charge is reduced with respect to the non-metals, which come from the right of the Periodic Table as we face it.

And thus metals, especially Group 1 and Group 2 metals, TEND to be reducing. It is very easy to form #Li^+#, and #Na^+#, and #Mg^(2+)# ions, i.e. alkali metal, and alkaline earth ions. Towards the right of the Periodic Table, nuclear charge has increased, and the incomplete valence shell SHIELDS the nuclear charge very ineffectively. The result is that as we go across a Period from left to right, the individual atoms become smaller, and ALSO, ionization energies become larger. This is a Periodic Trend with which you have to become very familiar.

The result? When a non-metal forms an ion, it tends to do so by REDUCTION, i.e. addition of an electron to the valence shell. Both oxygen and fluorine form anions, #O^(2-)#, and #F^-#, with a complete valence shell. Because the electron adds to the shell, and there is not a compensating increase in nuclear charge, the atomic radii of ANIONS should be much greater than that of the parent atom.

Anyway, I would look in your text for data: specifically ionic versus atomic radius for cations and anions and elements.