Animated Solution for Chemistry - Periodicity in Properties: Which one of the following ions has the highest value of ionic radius ?

A Quest for Size

Welcome to the microscopic arena of atoms and ions! In this thrilling problem, we are tasked with finding the giant among four distinct chemical species: the lithium ion (${\text{Li}}^{+}$), the boron ion (${\text{B}}^{3+}$), the oxide ion (${\text{O}}^{2-}$), and the fluoride ion (${\text{F}}^{-}$).

To solve this, we must dive deep into the quantum tug-of-war that dictates the size of an ion: the inward pull of the positively charged nucleus versus the outward push of the negatively charged electron cloud.

Anions: The First Cut

Before we look at the numbers, let's categorize our contenders. We have two cations (${\text{Li}}^{+}$ and ${\text{B}}^{3+}$) and two anions (${\text{O}}^{2-}$ and ${\text{F}}^{-}$).

When an atom loses electrons to become a cation, two things happen. First, it often loses its entire outermost electron shell. Second, the remaining electrons experience a much stronger effective nuclear charge because there are now more protons than electrons. This causes the electron cloud to shrink dramatically.

Conversely, when an atom gains electrons to become an anion, the added electrons increase the electron-electron repulsion within the valence shell. The nucleus, whose proton count hasn't changed, now struggles to hold onto the expanded electron cloud. Therefore, anions are generally much larger than cations of the same period. Right away, we can eliminate ${\text{Li}}^{+}$ and ${\text{B}}^{3+}$ from our search for the largest ion.

Now, let's focus on our heavyweights: the oxide ion (${\text{O}}^{2-}$) and the fluoride ion (${\text{F}}^{-}$).

Let's count their electrons. A neutral oxygen atom has an atomic number ($Z$) of 8, meaning it has 8 protons and 8 electrons. The $2-$ charge indicates it has gained 2 extra electrons, bringing its total to 10 electrons.

A neutral fluorine atom has an atomic number ($Z$) of 9, meaning 9 protons and 9 electrons. The $1-$ charge indicates it has gained 1 extra electron, also bringing its total to 10 electrons.

Because both ${\text{O}}^{2-}$ and ${\text{F}}^{-}$ possess exactly 10 electrons, they are called isoelectronic species. They both have the exact same electron configuration: $1s^{2}2s^{2}2p^6$, which is the stable configuration of the noble gas Neon.

Oxygen vs. Fluorine

If both ions have the exact same number of electrons arranged in the exact same way, what determines their size? The answer lies in the nucleus.

For isoelectronic species, the ionic radius is inversely proportional to the atomic number ($Z$). Mathematically, we write this as:

Why does this happen? Imagine the nucleus as a magnet pulling on the electron cloud. The oxide ion has 8 protons pulling on 10 electrons. The fluoride ion has 9 protons pulling on the same 10 electrons.

Because 9 protons exert a stronger electrostatic force than 8 protons, the nucleus of the fluoride ion pulls its electron cloud closer and tighter than the nucleus of the oxide ion. Consequently, the electron cloud of ${\text{O}}^{2-}$ is allowed to expand more freely.

Therefore, the oxide ion (${\text{O}}^{2-}$) emerges victorious as the ion with the highest value of ionic radius among the given options. It is a beautiful demonstration of how the delicate balance between nuclear charge and electron repulsion governs the architecture of the chemical world.