Question

The following is the correct increasing order of the ionic radii:

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

Answer 1

The correct option is B $A{l}^{3+}<L{i}^{+}<M{g}^{2+}<K^{+}$

${}_{3}Li{}_{12}Mg{}_{19}K{}_{13}Al$

atomic radii of neutral elements $Li<Mg<Al<K$

$L{i}^{+}M{g}^{2+}A{l}^{3+}{K}^{+}$

$2^{e-}{10}^{e-}{10}^{e-}{18}^{e-}$

$3p^{+}12p^{+}13p^{+}19p^{+}$

$\therefore$ ionic radii order,

$A{l}^{3+}<L{i}^{+}<M{g}^{2+}<K^{+}$

$L{i}^{+}$ is smaller than $M{g}^{2+}$ but larger than $A{l}^{3+}$ as

$ionicradii\propto \frac{1}{charge}$