Question

The correct sequence which shows the decreasing order of the ionic radii of the elements is:

• A. ${Al}^{3+}>M{g}^{2+}>N{a}^{+}>F>O^{2-}$
• B. $N{a}^{+}>M{g}^{2+}>{Al}^{3+}>O^{2-}>F^{-}$
• C. $N{a}^{+}>F^{-}>M{g}^{2+}>O^{2-}>{Al}^{3+}$
• D. $O^{2-}>F^{-}>N{a}^{+}>M{g}^{2+}>{Al}^{3+}$

Answer 1

The correct option is D $O^{2-}>F^{-}>N{a}^{+}>M{g}^{2+}>{Al}^{3+}$

For isoelectronic species, higher the $\frac{Z}{e}$ ratio, smaller is the ionic radius.
$O^{2-}=\frac{8}{10}=0.8$
$F^{-}=\frac{9}{10}=0.9$
$N{a}^{+}=\frac{11}{10}=1.1$
$M{g}^{2+}=\frac{12}{10}=1.2$
${Al}^{3+}=\frac{13}{10}=1.3$