Animated Solution for Chemistry - Periodicity in Properties: The size of the iso-electronic species ${\text{Cl}}^{-}$, $\text{Ar}$ and ${\text{Ca}}^{2+}$ is affected by

Imagine an atom as a microscopic game of tug-of-war. On one side, you have the positively charged protons sitting in the nucleus, pulling inward. On the other side, you have the negatively charged electrons orbiting in the electron cloud, trying to spread out due to their mutual repulsion. The final size of the atom or ion is simply the equilibrium point of this intense battle.

When we look at the species ${\text{Cl}}^{-}$, $\text{Ar}$, and ${\text{Ca}}^{2+}$, we are looking at a very special scenario. Let's count their electrons. A neutral Chlorine atom has $17$ electrons, but the ${\text{Cl}}^{-}$ ion has gained one, giving it $18$. Argon is a noble gas that naturally possesses $18$ electrons. Calcium normally has $20$ electrons, but the ${\text{Ca}}^{2+}$ ion has lost two, leaving it with exactly $18$ electrons.

Because they all have exactly $18$ electrons, we call them isoelectronic species (from the Greek iso meaning 'equal').

Nuclear Charge

Since all three species have the exact same number of electrons arranged in the exact same energy levels (up to the $n=3$ shell), the outward push—the electron-electron repulsion—is practically identical for all of them.

So, if the outward push is a tie, what decides the winner of the tug-of-war? The inward pull!

This inward pull is determined by the nuclear charge, which is simply the number of protons in the nucleus (the atomic number, $Z$). Let's look at the nuclei of our three contenders:

${\text{Cl}}^{-}$ has $17$ protons. $\text{Ar}$ has $18$ protons. * ${\text{Ca}}^{2+}$ has $20$ protons.

Even though ${\text{Ca}}^{2+}$ has the same number of electrons as the others, its nucleus is armed with $20$ protons. This massive positive charge exerts a tremendous electrostatic pull on the $18$ electrons, dragging the entire electron cloud closer to the center. Consequently, ${\text{Ca}}^{2+}$ shrinks and becomes the smallest of the three.

On the other end of the spectrum, ${\text{Cl}}^{-}$ only has $17$ protons trying to hold onto those same $18$ electrons. The inward pull is much weaker, allowing the electron cloud to expand outward, making ${\text{Cl}}^{-}$ the largest.

Mathematically, for isoelectronic species, the ionic radius ($r$) is inversely proportional to the atomic number ($Z$):

Therefore, the size of isoelectronic species is entirely dictated by the nuclear charge. The greater the nuclear charge, the smaller the ionic radius. This makes option (d) the perfect answer.