Chemistry - PERIODIC TABLE Question with Solution | TestHub

For isoelectronic species higher the $\frac{\mathrm{Z}}{\mathrm{e}}$ ratio, smaller the ionic radius $\frac{z}{e}$ for $$\begin{array}{rl}& {\mathrm{O}}^{2-}=\frac{8}{10}=0.8\\ & {\mathrm{F}}^{-}=\frac{9}{10}=0.9\\ & {\mathrm{Na}}^{+}=\frac{11}{10}=1.1\\ & {\mathrm{Mg}}^{2+}=\frac{12}{10}=1.2\\ & {\mathrm{Al}}^{3+}=\frac{13}{10}=1.3\end{array}$$