Q39 / 180 · KCET 2014 · Chemistry
The correct arrangement for the ions in the increasing order of their radii is
- Correct answerA\( \mathrm{Ca}^{2+}, \mathrm{K}^{+}, \mathrm{S}^{2-} \)
- B\( \mathrm{Ol}^{-}, \mathrm{F}^{-}, \mathrm{S}^{2-} \)
- C\( \mathrm{Na}^{+}, \mathrm{Cl}^{-}, \mathrm{Ca}^{2+} \)
- D\( N a^{+}, A l^{3+}, B e^{2+} \)
Solution
The number of electrons and the group number of the given species are given below:
$\begin{array}{|l|l|l|l|l|} \hline Ions & At. No & Group & Period & No. of e^{-} s \\ \hline S^{2-} & 16 & 16 & 3 & 18 \\ \hline K^{+} & 13 & 1 & 4 & 18 \\ \hline C a^{2+} & 20 & 2 & 4 & 18 \\ \hline \end{array}$
There ions are isoelectronic, therefore \( S^{2-} \) have greater size due to higher effective nuclear charge \( \left(Z_{\text {eff }}\right) \) followed by
\( \mathrm{K}^{+} \)then \( \mathrm{Ca}^{2+} \) Hence, the correct order of decreasing ionic radii is \( \mathrm{Ca}^{2+} < \mathrm{K}^{+} < \mathrm{S}^{2-} \)
$\begin{array}{|l|l|l|l|l|} \hline Ions & At. No & Group & Period & No. of e^{-} s \\ \hline S^{2-} & 16 & 16 & 3 & 18 \\ \hline K^{+} & 13 & 1 & 4 & 18 \\ \hline C a^{2+} & 20 & 2 & 4 & 18 \\ \hline \end{array}$
There ions are isoelectronic, therefore \( S^{2-} \) have greater size due to higher effective nuclear charge \( \left(Z_{\text {eff }}\right) \) followed by
\( \mathrm{K}^{+} \)then \( \mathrm{Ca}^{2+} \) Hence, the correct order of decreasing ionic radii is \( \mathrm{Ca}^{2+} < \mathrm{K}^{+} < \mathrm{S}^{2-} \)
Hence, the correct answer is option: 'A'.