$Cl ^{-}-1 s ^2 2 s ^2 2 p ^6 3 s ^2 3 p ^6$
It has 18 electrons and $17$ protons
Ar $-1 s^2 2 s^2 2 p^6 3 s^2 3 p^6$
It has 18 electrons and $18$ protons
$K ^{+}-1 s ^2 2 s ^2 2 p ^6 3 s ^2 3 p ^6$
It has 18 electrons and 19 protons
$Ca ^{2+}-1 s ^2 2 s ^2 2 p ^6 3 s ^2 3 p^6$
It has 18 electrons and $20$ protons
From this information, we can easily conclude that $Cl ^{-}$would be the largest because it has a higher number of electrons as compared to protons and so the 'effective nuclear charge' would be lower on each electron.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$Ca\left( s \right)|C{a^{ + 2}}_{\left( {aq.} \right)}||F{e^{ + 2}}_{\left( {aq} \right)}|Fe\left( s \right)$
$E_{C{a^{ + 2}}/Ca}^o = - 2.87\,V$ ; $E_{Fe/F{e^{ + 2}}}^o = 0.41\,V$