Question

The correct order of ionic radii of ions $N{a}^{+},M{g}^{2+},A{l}^{3+}$ and $S{i}^{4+}$ is

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

Answer 1

Answer: (D)

$N{a}^{+}>M{g}^{2+}>A{l}^{3+}>S{i}^{4+}$ . All are isoelectronic but nuclear charge per electron is greatest for $S{i}^{4+}$ . So it has smallest size and nuclear charge per electron for $N{a}^{+}$ is smallest. So it has largest size.