Animated Solution for Chemistry - Periodicity in Properties: The ionic radius of ${\text{Na}}^{+}$ ions is $1.02\text{A˚}$. The ionic radii (in $\text{A˚}$) of ${\text{Mg}}^{2+}$ and ${\text{Al}}^{3+}$, respectively, are

Meet the Ions

When we look at the periodic table, we often think of atoms as rigid spheres. But when atoms gain or lose electrons to become ions, their sizes change dramatically. In this problem, we are introduced to three specific cations: the sodium ion (${\text{Na}}^{+}$), the magnesium ion (${\text{Mg}}^{2+}$), and the aluminum ion (${\text{Al}}^{3+}$).

We are given a crucial piece of data: the ionic radius of ${\text{Na}}^{+}$ is exactly $1.02\text{A˚}$. Our mission is to logically deduce the radii of ${\text{Mg}}^{2+}$ and ${\text{Al}}^{3+}$ from the given multiple-choice options. To do this, we don't need to memorize a table of radii; we just need to understand the fundamental physics of the atom.

The first step in solving any problem involving ionic radii is to count the electrons. Let's break it down:

Sodium ($\text{Na}$) has an atomic number of 11. Losing one electron to form ${\text{Na}}^{+}$ leaves it with $11-1=10$ electrons. Magnesium ($\text{Mg}$) has an atomic number of 12. Losing two electrons to form ${\text{Mg}}^{2+}$ leaves it with $12-2=10$ electrons. Aluminum ($\text{Al}$)* has an atomic number of 13. Losing three electrons to form ${\text{Al}}^{3+}$ leaves it with $13-3=10$ electrons.

Notice a pattern? All three of these ions possess exactly 10 electrons. In chemistry, we call a group of atoms or ions with the exact same number of electrons isoelectronic species. Because they have the same number of electrons, their electron clouds have the same basic structural arrangement (the $1s^{2}2s^{2}2p^6$ configuration, identical to the noble gas Neon).

If the electron clouds are structurally identical, why would their sizes differ? The answer lies deep within the core of the atom: the nucleus.

While the number of electrons is constant at 10, the number of positively charged protons (the atomic number, $Z$) is different for each ion:

${\text{Na}}^{+}$ has $Z=11$ protons. ${\text{Mg}}^{2+}$ has $Z=12$ protons. * ${\text{Al}}^{3+}$ has $Z=13$ protons.

Imagine a microscopic tug-of-war. The positively charged protons in the nucleus are pulling inward on the negatively charged electrons. According to Coulomb's Law, a greater amount of positive charge will exert a stronger attractive force on the same number of electrons.

As we move from ${\text{Na}}^{+}$ to ${\text{Mg}}^{2+}$ to ${\text{Al}}^{3+}$, the nuclear charge increases from $+11$ to $+12$ to $+13$. This increasing positive charge pulls the 10 electrons closer and closer to the nucleus. Therefore, for isoelectronic species, the ionic radius is inversely proportional to the effective nuclear charge ($Z_{\text{eff}}$).

Based on our tug-of-war logic, the ion with the most protons will be the smallest, and the ion with the fewest protons will be the largest. This gives us a strict mathematical inequality for their sizes:

We are given that $r_{{\text{Na}}^{+}}=1.02\text{A˚}$. Substituting this into our inequality, we get:

Now, we simply inspect the given options to find the pair of numbers that are both smaller than $1.02$ and are in descending order.

(a) $1.05$ and $0.99$: Incorrect, $1.05$ is larger than $1.02$. (b) $0.72$ and $0.54$: Correct! Both are smaller than $1.02$, and $0.72>0.54$. (c) $0.85$ and $0.99$: Incorrect, the order is reversed ($0.85$ is not greater than $0.99$). (d) $0.68$ and $0.72$: Incorrect, the order is reversed.

The elegant physics of the nucleus leads us directly to the correct answer: Option (b).