Question

$N{a}^{+}.M{g}^{2+},A{l}^{3+}$ and $S{i}^{4+}$ are isoelectronic. The order of their ionic size is

• A. $N{a}^{+}>M{g}^{2+}<A{l}^{3+}<S{i}^{4+}$
• B. $N{a}^{+}<M{g}^{2+}>A{l}^{3+}>S{i}^{4+}$
• C. $N{a}^{+}>M{g}^{2+}>A{l}^{3+}>S{i}^{4+}$
• D. $N{a}^{+}<M{g}^{2+}>A{l}^{3+}<S{i}^{4+}$

Answer 1

The correct option is C

$N{a}^{+}>M{g}^{2+}>A{l}^{3+}>S{i}^{4+}$

In isoelectronic species the number of electrons are same but nuclear charge is different. As the nuclear charge increase, the attraction force on last electron increases, so the size decreases or in other words
$Ionicsize\propto \frac{1}{Chargeoncation}$ and hence, order is
$\stackrel{M{g}^{2+}>A{l}^{3+}>S{i}^{4+}}{\to }$
$\underset{Sizedecrease}{\overset{Nuclearchargeincrease}{\to }}$