The size of the isoelectronic species Cl-, Ar and Ca2+ is affected by:

🧠 In an isoelectronic set, the electron count, the electron arrangement, and the principal/azimuthal quantum numbers are all identical — only the proton count differs $\ce{Cl^-}$, Ar, $\ce{Ca^{2+}}$ all have 18 electrons in $[\ce{Ne}]\,3s^2\,3p^6$ configuration. So:

• The number of electrons is the same.
• The orbital arrangement (which orbitals are filled) is the same.
• The principal quantum number of the valence shell is $n = 3$ for all three.
• The azimuthal quantum numbers ($\ell$) of the highest-occupied subshell are also identical.

The only thing that differs across the three is the number of protons — i.e., nuclear charge $Z$.

🗺️ Eliminate options (b) Azimuthal quantum number — same for all three. Cannot affect size differences. ✗ (c) Electron-electron interactions in outer orbitals — same number of electrons in identical orbitals. The repulsion is identical too. ✗ (d) Principal quantum number — all three valence shells are $n = 3$. Same value. ✗ (a) Nuclear charge — Cl has $Z=17$, Ar has $Z=18$, Ca has $Z=20$. Different. This is the only varying factor, so this is what changes the size.

⚠️ The trap Option (c) sounds plausible because "more electrons → more repulsion → bigger". But in an isoelectronic set the electron count is by definition fixed. The repulsion doesn't change. Only the proton-count difference matters. When students don't see "isoelectronic" written boldly, they reach for electron-related answers reflexively. Always check the electron count first.

$\boxed{\text{Answer: (a)}}$