Question

The sizes of the following species in the increasing order can be given as:

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

Answer 1

The correct option is A ${Mg}^{2+}<{Na}^{+}<F^{-}<Al$

For isoelectronic ions, more the charge on cation, lesser is the ionic radius. For ${Mg}^{2+}(Z=12),{Na}^{+}(Z=11)$. $F^{-}(Z=9)$. Thus, $F^{-}$ has largest size among ${Mg}^{2+},{Na}^{+},F^{-}$ becuase it is anion. More the charge on anion, more the radius.
Now $Al$ has higher radius than $F^{-}$. Hence the increasing order of the size is ${Mg}^{2+}<{Na}^{+}<F^{-}<Al$