Question

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

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

Answer 1

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

In case of isoelectronic species, the greater the nuclear charge the greater is the attraction for electrons and smaller is the ionic radius.