Question
The correct order of the decreasing ionic radii among the following isoelectronic species is
A.
$$C{a^{2 + }} > {K^ + } > {S^{2 - }} > C{l^ - }$$
B.
$$C{l^ - } > {S^{2 - }} > C{a^{2 + }} > {K^ + }$$
C.
$${S^{2 - }} > C{l^ - } > {K^ + } > C{a^{2 + }}$$
D.
$${K^ + } > C{a^{2 + }} > C{l^ - } > {S^{2 - }}$$
Answer :
$${S^{2 - }} > C{l^ - } > {K^ + } > C{a^{2 + }}$$
Solution :
Key Idea Ionic radii $$ \propto $$ charge on anion $$ \propto \frac{1}{{{\text{charge}}\,{\text{on}}\,{\text{cation}}}}$$
During the formation of a cation, the electrons are lost from the outer shell and the remaining electrons experience a great force of attraction by the nucleus, ie. attracted more towards the nucleus. In other words, nucleus hold the remaining electrons more tightly and this results in decreased radii.
However, in case of anion formation, the addition of electron(s) takes place in the same outer shell, thus the hold of nucleus on the electrons of outer shell decreases and this results in increased ionic radii.
Thus, the correct order of ionic radii is $${S^{2 - }} > C{l^ - } > {K^ + } > C{a^{2 + }}$$