- ✓$C{l^ - }$
- 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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if $T_1 < T_2$ then $X$ and $Y$ respectively are
$(i)$ $C{H_3}\mathop {\mathop C\limits_{||} }\limits_O - {O^ - }$ $(ii)$ $C{H_3}{O^ - }$
$(iii)$ $C{N^ - }$ $(iv)$ $Image$
is