Question

The size of the following species increases in the order:

• A. $M{g}^{2+}<N{a}^{+}<F^{-}<Al$
• B. $F^{-}<Al<N{a}^{+}>M{g}^{2+}$
• C. $Al<M{g}^{2+}<F^{-}<N{a}^{+}$
• D. $N{a}^{+}<Al<F^{-}<M{g}^{2+}$

Answer 1

The correct option is A $M{g}^{2+}<N{a}^{+}<F^{-}<Al$

No. of electrons in $M{g}^{2+}$ = $10$

No. of electrons in $N{a}^{+}$ =$10$

No. of electrons in $F^{-}$ = $10$

No. of electrons in $Al=13$

Effective Nuclear charge = {$\frac{nuclearcharge}{No.ofelectrons}$ } experienced by$Al$ atom is lowest ,thus it has largest size.

Among the isoelectronic ions, the trend is $F^{-}>N{a}^{+}>M{g}^{2+}$ which can be explained as; size increases as we move down a group and decreases as we move from left to right across a period.

Hence, option A is correct.