How do you determine if an ion is larger?

For example, which ion is larger in each pair?
#Ca^(2+)#, #Mg^(2+)#
#Cl^-#, #P^(3-)#
#Cu^+#, #Cu^(2+)#

1 Answer
anor277
Dec 4, 2016

#"A priori"#, the more positive the cation, the SMALLER should be the ion...........

Explanation:

When a cation forms, it must necessarily lose an electron. The electron is removed from the valence shell, and is no longer available to shield the nuclear charge. The cationic radius should thus contract with respect to the atomic radius.

On the other hand, when an anion forms, a given atom is reduced, and accepts electrons into its valence shell. The valence should expand, and thus anions should have a larger radius than their parent atoms.

For the alkaline earth cations, #Mg^(2+)#, and #Ca^(2+)#, you have a third Period cation versus a fourth Period cation. Clearly, the calcium ion should be larger! (Of course the ionic radius is reduced with respect to the parent atom, but this is not what we were asked to consider!).

For non-metal ions, #Cl^-#, and #P^(3-)#, you have third row anions. The more negative phosphide should be larger, in that (i) it has a reduced nuclear charge, i.e. #Z# is smaller, and (ii) it is trinegative rather than singly negative. If we were to consider the neutral atoms, certainly #Cl# would be the smaller atom. Why?

For #"cuprous"#, #Cu^+#, and #"cupric"# ions, #Cu^(2+)#, the dication should be SMALLER, as explained in the last section.

To end, you are a chemist, a physical scientist. We have considered some of the possible effects of ionic charge. It is up to you to consult the data, and find some actual measurements of ionic radii to see that we haven't got it #"backasswards"#. Here is a start, but you must consider the data.

Related questions
See all questions in Periodic Trends in Ionic Size
Impact of this question
37776 views around the world
You can reuse this answer
Creative Commons License