Question

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

• A. $Mg$ atom has less no. of neutrons than $Al$.
• B. $A{l}^{+3}$ has a higher nuclear charge than $M{g}^{+2}$
• C. $A{l}^{+3}$ has a lesser nuclear charge than $M{g}^{+2}$
• D. electropositivity increases from left to right in a period.

Answer 1

Answer: (B)

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

Higher the electrostatic attraction between nucleus and valence electron, smaller will be the size of the atom/ion.
Electrostatic attraction is given by:
$Zeff=Z-S$
where Z= the number of protons in the nucleus of an atom or ion (the atomic number), and
S= shielding from core electrons.
In case of $A{l}^{3+}$ and $M{g}^{2+}$, they are isoelectronic species and thus have same number of electrons as $1s^{2}2s^{2}2p^6$
isoelectric series of atoms and ions with differing numbers of protons (and thus different nuclear attraction), gives the relative ionic sizes of each atom or ion with respect to atomic number.
Atomic number, $Z_{Al}=13$ and $Z_{Mg}=12$. Thus $A{l}^{3+}$ has lower ionic radii than $M{g}^{2+}$ due to higher nuclear charge.
Option B is correct