- ✓$C{l^ - }$
- B$Ar$
- C${K^ + }$
- D$C{a^{2 + }}$
$Cl ^{-}-1 s ^2 2 s ^2 2 p ^6 3 s ^2 3 p ^6$
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.
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Reason : Chlorination of $CH_4$ takes place in sunlight.


$\left. \begin{gathered}
{\text{4N}}{{\text{H}}_{3(g)}}{\text{ + 5}}{{\text{O}}_{2(g)}} \to {\text{ 4N}}{{\text{O}}_{(g)}}{\text{ + 6}}{{\text{H}}_{\text{2}}}{{\text{O}}_{{\text{(l)}}}} \hfill \\
{\text{2N}}{{\text{O}}_{(g)}}{\text{ + }}{{\text{O}}_{2(g)}} \to {\text{2N}}{{\text{O}}_{2(g)}} \hfill \\
{\text{3N}}{{\text{O}}_{2(g)}}{\text{ + }}{{\text{H}}_{{\text{ 2}}}}{{\text{O}}_{(l)}} \to {\text{2H N}}{{\text{O}}_{3(aq)}}{\text{ + N}}{{\text{O}}_{(g)}} \hfill \\
\end{gathered} \right|\begin{array}{*{20}{c}}
{\Delta H(kJ)} \\
{ - 904} \\
{ - 112} \\
{ - 140}
\end{array}$