Question

The ionic radii of $K^{+},N{a}^{+},A{l}^{3+}\text{and}M{g}^{2+}$ are in the order

• A. $N{a}^{+}<K^{+}<M{g}^{2+}<A{l}^{3+}$
• B. $A{l}^{3+}<M{g}^{2+}<K^{+}<N{a}^{+}$
• C. $A{l}^{3+}<M{g}^{2+}<N{a}^{+}<K^{+}$
• D. $K^{+}<A{l}^{3+}<M{g}^{2+}<N{a}^{+}$

Answer 1

The correct option is C $A{l}^{3+}<M{g}^{2+}<N{a}^{+}<K^{+}$

	Ionic radii (in pm)
$N{a}^{+}$	$102$
$K^{+}$	$138$
$M{g}^{2+}$	$72$
$A{l}^{3+}$	$53.5$

Al, Mg and Na are in third period.
Generally among the ions of same period, higher the charge on cation smaller will be its ionic radius.
Potassium belongs to the fourth period and shall have highest radius than the rest.
$A{l}^{3+}<M{g}^{2+}<N{a}^{+}<K^{+}$
Option (c) is the correct answer.