- ✓$C{l^ - }$
- 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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$(1)$ $\begin{array}{*{20}{c}}
{C{H_3}CH = C - C{H_3}} \\
{\,\,\,\,\,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}}
\end{array}$
$(2)$ $C{H_3}CH = CHC{H_3}$
$(3)$ $\mathop {C{H_3}C}\limits^{\begin{subarray}{l}
\,\,\,\,\,\,\,\,{\begin{array}{*{20}{c}}
{}&H
\end{array}} \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,|
\end{subarray} } = CHC{H_2}C{H_3}$
$(4)$ $\begin{array}{*{20}{c}}
{C{H_3}C = C - C{H_3}} \\
{\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_{3\,\,}}\,\,C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$