Animated Solution for Chemistry - Periodicity in Properties: The correct order of the ionic radii of ${\text{O}}^{2-}$, ${\text{N}}^{3-}$, ${\text{F}}^{-}$, ${\text{Mg}}^{2+}$, ${\text{Na}}^{+}$ and ${\text{Al}}^{3+}$ is

Understanding Isoelectronic Species

Imagine a microscopic tug-of-war happening inside every atom and ion. On one side, you have the positively charged protons in the nucleus pulling inward. On the other side, you have the negatively charged electrons in the outer shells trying to spread out due to their mutual repulsion. The balance of this tug-of-war determines the ultimate size, or ionic radius, of the species.

In this problem, we are given a set of six ions: ${\text{O}}^{2-}$, ${\text{N}}^{3-}$, ${\text{F}}^{-}$, ${\text{Mg}}^{2+}$, ${\text{Na}}^{+}$, and ${\text{Al}}^{3+}$. At first glance, they look completely different. They belong to different groups and periods. But if we look closely and count their electrons, a beautiful pattern emerges.

Let's do the math: - Nitrogen ($Z=7$) gains 3 electrons: $7+3=10$ electrons. - Oxygen ($Z=8$) gains 2 electrons: $8+2=10$ electrons. - Fluorine ($Z=9$) gains 1 electron: $9+1=10$ electrons. - Sodium ($Z=11$) loses 1 electron: $11-1=10$ electrons. - Magnesium ($Z=12$) loses 2 electrons: $12-2=10$ electrons. - Aluminum ($Z=13$) loses 3 electrons: $13-3=10$ electrons.

Every single one of these ions has exactly 10 electrons. In chemistry, species that possess the exact same number of electrons are called isoelectronic species.

Now, if they all have the same number of electrons, shouldn't they all be the same size? Absolutely not! This is where our tug-of-war analogy comes back into play. While the number of electrons (the outward force) is constant across all these ions, the number of protons (the inward pulling force) is drastically different.

The inward pull exerted by the nucleus on the outermost electrons is known as the effective nuclear charge ($Z_{\text{eff}}$). For isoelectronic species, the shielding effect caused by inner electrons is roughly the same. Therefore, the effective nuclear charge is directly proportional to the actual nuclear charge, which is simply the atomic number $Z$.

Mathematically, we can state that for isoelectronic species:

Let's line up our competitors based on their atomic numbers (number of protons): - ${\text{N}}^{3-}$ has $Z=7$ - ${\text{O}}^{2-}$ has $Z=8$ - ${\text{F}}^{-}$ has $Z=9$ - ${\text{Na}}^{+}$ has $Z=11$ - ${\text{Mg}}^{2+}$ has $Z=12$ - ${\text{Al}}^{3+}$ has $Z=13$

Aluminum has a whopping 13 protons pulling on just 10 electrons. This massive inward force shrinks the electron cloud significantly, making ${\text{Al}}^{3+}$ the smallest ion in the group. Conversely, Nitrogen only has 7 protons trying to hold onto 10 electrons. The nucleus struggles to keep them close, allowing the electron cloud to expand, making ${\text{N}}^{3-}$ the largest.

Therefore, as the atomic number $Z$ increases, the ionic radius decreases. Arranging them in increasing order of their size gives us:

This perfectly matches option (c). Whenever you face a question about ionic radii in JEE, your first instinct should always be to check if the species are isoelectronic. If they are, simply count the protons!