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

Answer: (C)

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
Ionic size $\propto \frac{1}{Chargeoncation}$ - and hence, order is
$\stackrel{N{a}^{+}>M{g}^{2+}>A{I}^{3+}>S{i}^{4+}}{\to }$
$\underset{Sizedecrease}{\overset{Nuclearchargeincrease}{\to }}$