Question

The correct order of ionic size of $N^{3-}$, $N{a}^{+}$, $F^{-}$, $M{g}^{2+}$ and $O^{2-}$ is:

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

Answer 1

Answer: (C)

$M{g}^{2+}<N{a}^{+}<F^{-}<O^{2-}<N^{3-}$

The correct order of ionic size is $M{g}^{2+}<N{a}^{+}<F^{-}<O^{2-}<N^{3-}$

These are isoelectronic species. They have same number of electrons (10 electrons in each ion).

$M{g}^{2+}$ has maximum number of protons (12). It has maximum attraction between nucleus and valence electrons as more number of protons act on given number of electrons. Hence, valence electrons are pulled strongly, which results in lowest ionic size.
$N^{3-}$ has minimum number of protons (7). It has minimum attraction between nucleus and valence electrons as less number of protons act on given number of electrons. Hence, valence electrons are pulled least strongly, which results in highest ionic size.