Animated Solution for Chemistry - Periodicity in Properties: The ionic radii (in $\text{A˚}$) of ${\text{N}}^{3-}$, ${\text{O}}^{2-}$ and ${\text{F}}^{-}$ respectively are

Understanding Isoelectronic Radii

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 can change dramatically. In this problem, we are tasked with finding the correct order of ionic radii for three specific ions: ${\text{N}}^{3-}$, ${\text{O}}^{2-}$, and ${\text{F}}^{-}$.

To solve this, we need to look beyond just the symbols and dive into the subatomic particles that dictate their physical size.

The first step in analyzing any group of ions is to count their electrons. Let's break it down:

Nitrogen ($\text{N}$) has an atomic number of $7$. The $3-$ charge means it has gained $3$ extra electrons. Total electrons = $7+3=10$. Oxygen ($\text{O}$) has an atomic number of $8$. The $2-$ charge means it has gained $2$ extra electrons. Total electrons = $8+2=10$. Fluorine ($\text{F}$)* has an atomic number of $9$. The $1-$ charge means it has gained $1$ extra electron. Total electrons = $9+1=10$.

Notice a pattern? All three ions have exactly $10$ electrons. In chemistry, species with the exact same number of electrons are called isoelectronic species. Because their electron clouds contain the same number of electrons, the repulsion between these electrons is roughly similar across all three ions.

If the electron clouds are similar, why would their sizes be different? The answer lies in the nucleus.

Imagine a tug of war between the positively charged protons in the nucleus and the negatively charged electrons in the cloud.

In ${\text{N}}^{3-}$, there are only $7$ protons pulling on $10$ electrons. This is a relatively weak pull, allowing the electron cloud to expand outward. In ${\text{O}}^{2-}$, there are $8$ protons pulling on the same $10$ electrons. The pull is stronger, drawing the cloud closer. In ${\text{F}}^{-}$, there are $9$ protons* pulling on the $10$ electrons. This is the strongest pull of the three, compressing the electron cloud the most.

This concept is governed by the Effective Nuclear Charge ($Z_{\text{eff}}$). For isoelectronic species, as the atomic number ($Z$) increases, the effective nuclear charge increases, and the ionic radius strictly decreases.

Mathematically, we can say:

Based on our logic, the order of the number of protons is $Z_{\text{N}}<Z_{\text{O}}<Z_{\text{F}}$. Therefore, the order of their ionic radii must be the exact opposite:

Now, we simply look at the given options to find a sequence of numbers that strictly decreases.

Option (c) provides the values $1.71\text{A˚}$, $1.40\text{A˚}$, and $1.36\text{A˚}$. This perfectly matches our derived trend ($1.71>1.40>1.36$). Thus, Option (c) is the undeniably correct answer.

Always remember: when electrons are tied, the protons decide the size!