Question

The sizes of following species increase in the order:

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

Answer 1

The correct option is A $M{g}^{2+}$ $<$ $N{a}^{+}$ $<$ $F^{-}$ $<$ Al

For isoelectronic species, $M{g}^{2+}$, $N{a}^{+}$, $F^{-}$ the sizes increase in the order.
$M{g}^{2+}$ $<$ $N{a}^{+}$ $<$ $F^{-}$ $<$ Al

It is easier to deduce $M{g}^{2+}$ $<$ $N{a}^{+}$ $<$ $F^{-}$ because in the first 2 there is a loss of an entire shell and in the third, there is an addition of an electron keeping the nuclear charge the same.

It gets slightly trickier with $F^{-}$ and Al.The atomic radius of Al (143pm) is greater than ionic radius of $F^{-}$(136 pm). Instead of remembering this data, we follow the general rule that the addition of a new shell usually outweighs the addition of protons. So, the size of Al would be larger..