The size of the iso-electronic species Cl$$^{-}$$, Ar and Ca$$^{2+}$$ is affected by:

We begin by recalling that the term “iso-electronic species’’ refers to atoms or ions that possess exactly the same number of electrons. In our problem we have the chloride ion $$\text{Cl}^{-}$$, the neutral argon atom $$\text{Ar}$$ and the calcium dication $$\text{Ca}^{2+}$$. Each of these species contains $$18$$ electrons in total, so the electronic configuration of all three is

$$1s^{2}\;2s^{2}\;2p^{6}\;3s^{2}\;3p^{6}.$$

Because the electron count and the distribution among the shells and subshells are identical, the principal quantum number $$n$$ for the outermost electrons is the same ($$n = 3$$) and the azimuthal quantum numbers $$l$$ available in that shell (namely $$l = 0$$ for the $$3s$$ orbital and $$l = 1$$ for the $$3p$$ orbitals) are also the same for all three species. Hence, any difference in their ionic or atomic radii cannot arise from a change in either $$n$$ or $$l$$, because those quantum numbers are fixed across the series. Likewise, since the total number of electrons is the same, the magnitude of electron-electron repulsions in equivalent orbitals is essentially similar.

The factor that does differ is the actual number of protons, i.e., the nuclear charge $$Z$$. We list the values:

$$\text{For }\text{Cl}^{-}\!: Z = 17, \qquad \text{For }\text{Ar}\!: Z = 18, \qquad \text{For }\text{Ca}^{2+}\!: Z = 20.$$

The effective nuclear charge experienced by an electron is commonly represented by

$$Z_{\text{eff}} = Z - S,$$

where $$S$$ is the screening constant accounting for shielding by the other electrons. Since $$S$$ depends mainly on how many inner electrons are present—and here that number is the same for all three species—the variation in $$Z_{\text{eff}}$$ follows directly from the variation in $$Z$$.

As $$Z$$ increases from $$17$$ to $$18$$ to $$20$$, $$Z_{\text{eff}}$$ correspondingly increases. A higher effective nuclear charge pulls the common set of $$18$$ electrons closer to the nucleus, thereby reducing the ionic/atomic radius. Consequently we predict the size order

$$r\!\left(\text{Cl}^{-}\right) > r\!\left(\text{Ar}\right) > r\!\left(\text{Ca}^{2+}\right).$$

This trend is governed solely by the variation in nuclear charge. None of the other listed factors—azimuthal quantum number of the valence shell, electron-electron interactions (which remain comparable), or principal quantum number of the valence shell—changes among these iso-electronic species.

Hence, the factor affecting their relative sizes is the nuclear charge.

Hence, the correct answer is Option A.