Question

Select The Correct Order Of Ionic Radii For${\mathrm{Na}}^{+},{\mathrm{Mg}}^{2+},{\mathrm{F}}^{-}\mathrm{and}{\mathrm{O}}^{2-}$?

• A. ${\mathrm{Mg}}^{2+}<{\mathrm{Na}}^{+}<{\mathrm{F}}^{-}<{\mathrm{O}}^{2-}$
• B. ${\mathrm{Na}}^{+}<{\mathrm{Mg}}^{2+}<{\mathrm{F}}^{-}<{\mathrm{O}}^{2-}$
• C. ${\mathrm{O}}^{2-}<{\mathrm{F}}^{-}<{\mathrm{Na}}^{+}<{\mathrm{Mg}}^{2+}$
• D. ${\mathrm{Mg}}^{2+}<{\mathrm{O}}^{2-}<{\mathrm{Na}}^{+}<{\mathrm{F}}^{-}$

Answer 1

The correct option is A

${\mathrm{Mg}}^{2+}<{\mathrm{Na}}^{+}<{\mathrm{F}}^{-}<{\mathrm{O}}^{2-}$

Explanation for the correct option:

option(A):${\mathrm{Mg}}^{2+}<{\mathrm{Na}}^{+}<{\mathrm{F}}^{-}<{\mathrm{O}}^{2-}$

• The given compounds are basically isoelectronic in nature.
• Isoelectronic species are known as those elements which have the same number of electrons.
• The size of these isoelectronic species can be compared with the help of the relative number of protons present in the species.
• The species having more numbers in the nucleus tend to have more nuclear attraction towards the outermost electrons, hence the size of the species decreases accordingly.
• Among the given species, ${\mathrm{Mg}}^{2+},{\mathrm{Na}}^{+},{\mathrm{F}}^{-},{\mathrm{O}}^{2-}$ have $12,11,9\mathrm{and}8$ protons respectively in their outermost shells.
• Thus, ${\mathrm{Mg}}^{2+}$will have the smallest size and ${\mathrm{O}}^{2-}$ will have the largest size among the given ions.
• Hence, the correct order of the ionic radii for the given ion is :${\mathrm{Mg}}^{2+}<{\mathrm{Na}}^{+}<{\mathrm{F}}^{-}<{\mathrm{O}}^{2-}$.

Hence, the Correct Order Of Ionic Radii For${\mathrm{Na}}^{+},{\mathrm{Mg}}^{2+},{\mathrm{F}}^{-}\mathrm{and}{\mathrm{O}}^{2-}$is ${\mathrm{Mg}}^{2+}<{\mathrm{Na}}^{+}<{\mathrm{F}}^{-}<{\mathrm{O}}^{2-}$.