Chapter 12, Problem 61A

Interpretation Introduction

Interpretation:

The given ions are to arrange in increasing order of radius.

Concept introduction:

Atomic size depends on the placement of molecules in periodic table.

Left to right, atomic radius decreases and top to bottom, radius increases.

Same trend is applicable for ions.

Answer to Problem 61A

Increasing order of radius is:

  $M{g}^{2+}<N{a}^{+}<F^{-}<O^{2-}$

Explanation of Solution

The ions which are to be arranged in increasing order of radius are isoelectronic.

Ions	$M{g}^{2+}$	$N{a}^{+}$	$F^{-}$	$O^{2-}$
Protons	12	11	9	8
Electrons	10	10	10	10

Thus, all ions have same number of electrons but different number of protons.

As seen from table, effective nuclear charge of $M{g}^{2+}$ is maximum so it has smallest radius and effective nuclear charge of $O^{2-}$ is minimum so it has largest radius

So the trend is:

  $M{g}^{2+}<N{a}^{+}<F^{-}<O^{2-}$

Conclusion

Thus, in isoelectronic species, atomic radius increases with decrease in effective nuclear charge