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# The order of size is: A.${S^{2 - }}$>$C{l^ - }$>${O^{2 - }}$>${F^ - }$B.$C{l^ - }$>${S^{2 - }}$>${O^{2 - }}$>${F^ - }$C.${S^{2 - }}$>${O^{2 - }}$>$C{l^ - }$>${F^ - }$D.${S^{2 - }}$>${O^{2 - }}$>${F^ - }$>$C{l^ - }$  Verified
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Hint:
The order of the size of the given ions is an iteration of the period patterns that can be observed across the periodic table. These periodic trends arise from the electronic configurations of the elements associated that are parent to these ions.

The ions given to us are ${S^{2 - }}$,$C{l^ - }$,${O^{2 - }}$ and ${F^ - }$
Out of this, ${S^{2 - }}$and $C{l^ - }$are in the same period and the atomic number of sulphur is 16 and that of chlorine is 17 Hence the size of${S^{2 - }}$is greater than$C{l^ - }$.
Also${O^{2 - }}$ and ${F^ - }$are in the same period and the atomic number of oxygen is 8 and that of fluorine is 9 Hence the size of ${F^ - }$is greater than ${O^{2 - }}$
Also, ${O^{2 - }}$ and ${F^ - }$ belong to period 2 while ${S^{2 - }}$and $C{l^ - }$belong to period 3. Hence the relation between all their atomic sizes can be determined as:
${S^{2 - }}$>$C{l^ - }$>${O^{2 - }}$>${F^ - }$