The ionic radii of

Question:

The ionic radii of $\mathrm{K}^{+}, \mathrm{Na}^{+}, \mathrm{Al}^{3+}$ and $\mathrm{Mg}^{2+}$ are in the order:

  1. $\mathrm{Na}^{+}<\mathrm{K}^{+}<\mathrm{Mg}^{2+}<\mathrm{Al}^{3+}$

  2. $\mathrm{Al}^{3+}<\mathrm{Mg}^{2+}<\mathrm{K}^{+}<\mathrm{Na}^{+}$

  3. $\mathrm{Al}^{3+}<\mathrm{Mg}^{2+}<\mathrm{Na}^{+}<\mathrm{K}^{+}$

  4. $\mathrm{K}^{+}<\mathrm{Al}^{3+}<\mathrm{Mg}^{2+}<\mathrm{Na}^{+}$


Correct Option: , 3

Solution:

$\mathrm{Al}^{3+}, \mathrm{Mg}^{2+}$ and $\mathrm{Na}^{+}$are isoelectronic ionic species.

For monoatomic ionic isoelectronic species as positive charge increases ionic size decreases.

The order of size of $\mathrm{Na}^{+} \& \mathrm{~K}^{+}$is $\mathrm{Na}^{+}<\mathrm{K}^{+}$,

$\therefore$ order of ionic radii is : $\mathrm{Al}^{3+}<\mathrm{Mg}^{2+}<\mathrm{Na}^{+}<\mathrm{K}^{+}$

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