Periodicity II: the ionic radius

Similar to what we have known for atomic radius, the ionic radii of the most common ions of elements are following certain periodic trends as well.

We'll discuss the periodic trends of ionic radii for those elements in the main group, and we'll look at their most common ions with noble gas electron configurations. Once you understand the principles, you can compare any two ions easily.

Two factors

There are two factors that we must consider when we compare the ionic radii.

Energy levels

The first factor is the number of energy levels. Similar to comparing atomic radii, ionic radii are greatly determined by the number of energy levels.

More energy levels give rise to a larger ionic radius.

Charge of particle

The second factor is the charges of the ions. We could simply put the more positive the ion is, the smaller the ionic radius is; and the more negative the ion is, the larger the ionic radius is.

The rationale is that positvely charged cations will have greater attractions towards their electrons and therefore, show smaller radii. On the other hand, negatively charged anions would exert repulsion forces on the electrons, hence larger radii.

Ions of the same element

Now we'll use the two factors to compare ionic radii of several ions of the same element, for example, \( \ce{Fe^{2+}} \) and \( \ce{Fe^{3+}} \). To make things more interesting, we'll include Fe in the comparison as well.

The first thing that you need to do before comparing any radius is always writing the electron configurations. So let's do that.

Particle	Electron configuration	Energy levels	Charge
Fe	\( \ce{[Ar] 4s^{2} 3d^6} \)	4	0
\( \ce{Fe^{2+}} \)	\( \ce{[Ar] 3d^6} \)	3	+2
\( \ce{Fe^{3+}} \)	\( \ce{[Ar] 3d^5} \)	3	+3

From the table above we could see that, the atom has 4 energy levels, which give rise to the largest radius of the particle. On the other hand, both ions have the same number of energy levels (3), we would look at their charges instead. Since \( \ce{Fe^{2+}} \) has a lower positive charge (+2), it would have a larger radius as the electrons would experience less attraction towards the nucleus. Therefore, we could rank the radii of the three particles as \( \ce{Fe} \gt \ce{Fe^{2+}} \gt \ce{Fe^{3+}} \).

As you can see, we used the two factors (number energy levels and charge) to compare the radii. Can you do the comparison for the following particles and give reasons for your answers?

1. Cu, \( \ce{Cu^+} \), \( \ce{Cu^{2+}} \)
2. Mg, \( \ce{Mg^+} \), \( \ce{Mg^{2+}} \)
3. N, \( \ce{N^-} \), \( \ce{N^{2-}} \), \( \ce{N^{3-}} \)
4. Cl, \( \ce{Cl^-} \), \( \ce{Cl^{2-}} \)

Periodic trends of ionic radius

Down the group

It is fairly simple to compare the ionic radii of elements in the same group. The ions would have the same charge because they would gain or lose the same number of electrons in order to attain a noble gas electron configuration. For example, all common ions of alkali metals are having a +1 charge.

However, the number of energy levels increases down the group. We have listed down the numbers of energy levels together with the electron configurations of the common ions of alkali metals in the table below.

Ion	Electron configuration	Number of energy levels
\( \ce{Li^+} \)	\( \ce{1s^2} \)	1
\( \ce{Na^+} \)	\( \ce{1s^{2} 2s^{2} 2p^6} \)	2
\( \ce{K^+} \)	\( \ce{1s^{2} 2s^2 2p^6 3s^2 3p^6} \)	3
\( \ce{Rb^+} \)	\( \ce{1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6} \)	4

Therefore, we could simply conclude that the ionic radii of the common ions of elements increase down a group.

Across the period

It is a little more complicated when we discuss the periodic trend of ionic radii across a period. This is because the numbers of energy levels and charges of the common ions vary. However, we'll apply the same technique by comparing the energy levels and charges, and we can easily figure out the answers.

Let's use period 2 as our example and you can try period 3 by yourself.

Element	Li	Be	B	N	O	F
Ion	\( \ce{Li^+} \)	\( \ce{Be^{2+}} \)	\( \ce{B^{3+}} \)	\( \ce{N^{3-}} \)	\( \ce{O^{2-}} \)	\( \ce{F^{-}} \)
Electron configuration	\( \ce{1s^2} \)	\( \ce{1s^2} \)	\( \ce{1s^2} \)	\( \ce{1s^2 2s^2 2p^6} \)	\( \ce{1s^2 2s^2 2p^6} \)	\( \ce{1s^2 2s^2 2p^6} \)
Energy levels	1	1	1	2	2	2
Charge	+1	+2	+3	-3	-2	-1

We notice that all anions are having 1 more energy level as compared to those cations. This tells us that anions have greater ionic radii than cations.

If we focus on cations alone, they all have the same number of energy levels (1), but their charges are increasingly positive as we move across period 2. Hence, their ionic radii are decreasing across the period.

When we move on to the anions, their charges are becoming less negative across the period, while they have the same number of energy levels. Therefore, their ionic radii are decreasing across the period too.

In summary, we can say that the ionic radii of the common ions of elements decrease across a period, and radii of anions are larger than those of cations. We will show a simple illustration of the ionic radii below (not to scale).

Anions and cations in two periods

Always remember, we consider two factors when comparing ionic radii:

1. number of energy levels;
2. charge of ions.

So, we are able to compare ions in any position of the periodic table, not only in the same period or group.

Can you sort the following elements in a sequence of increasing ionic radii of the common ions:

P, S, Cl, K, Ca ? (Hint: construct a table similar to the one we showed above)

We'll show the answers to all these questions in the next post. Try them before you move on to the next page.