A) \[C{{s}^{+}}\]
B) \[L{{i}^{+}}\]
C) \[N{{a}^{+}}\]
D) \[{{K}^{+}}\]
Correct Answer: A
Solution :
In a group from top to bottom atomic size continuously increases because 1. number of shells (n) increases 2. Atomic number increases but effective nuclear charge \[({{Z}^{*}})\] almost remains constant due to screening. Since, \[{{r}_{n}}\propto \frac{{{n}^{2}}}{{{Z}^{*}}}\] \[{{r}_{n}}\propto {{n}^{2}}\] (\[{{Z}^{*}}\]remaining constant) The order of ionic radius in \[C{{s}^{+}},L{{i}^{+}},N{{a}^{+}}\] and \[{{K}^{+}}\] as \[C{{s}^{+}}>{{K}^{+}}>N{{a}^{+}}>L{{i}^{+}}\] Therefore, \[C{{s}^{+}}\] has largest ionic radius among given IA group ions.
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