Question

$A{l}^{3+}$ has a lower ionic radius than $M{g}^{2+}$ because:

• A. $A{l}^{3+}$ has a higher nuclear charge than $M{g}^{2+}$
• B. $Mg$ has lesser number of neutrons than $Al$
• C. $Mg$ and $Al$ differ in their electronegativity values
• D. $Al$ has a lower ionisation potential than $Mg$

Answer 1

Answer: (A)

$A{l}^{3+}$ has a higher nuclear charge than $M{g}^{2+}$

$A{l}^{3+}$ and $M{g}^{2+}$ have equal number of electrons and are, therefore, iso-electronic.

For iso-electronic species, size depends upon the effective nuclear charge and size is inversely proportional to the effective nuclear charge. More is the positive charge, more is the effective nuclear charge and thus, lesser is the size.

Since the effective nuclear charge is greater in $A{l}^{3+}$ than $M{g}^{2+}$, it has a smaller size than $M{g}^{2+}$.