Question

Assertion (A) : $M{g}^{2+}$ and $A{l}^{3+}$ are isoelectronic but the magnitude o $f$ ionic radius of $A{l}^{3+}$ is less than that in $M{g}^{2+}$
Reason (R ): The effective nuclear charge oil the outermost electrons in $A{l}^{+3}$ is greater than that in $M{g}^{2+}$

• A. (A) is true, (R) is true and (R) is the correct explanation for (A)
• B. (A) is true. (R) is true but (R) is the not the collect explanation for A
• C. (A) is true but (R) is false
• D. (A) is false but (R) is true

Answer 1

Answer: (A)

Higher the electrostatic attraction between nucleus and valence electron,
smaller will be the size of the atomVion.
Electrostatic attraction is given by $Z_{\text{eff }}=Z-S$
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 different 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.
Hence, $A$ is true, $R$ is true and $R$ is the correct explanation for $A$.