Question

Consider the following species: $N^{3–},O^{2–},F^{–},N{a}^{+},M{g}^{2+}\text{and}A{l}^{3+}$

$\left(B\right)$ Arrange them in the order of increasing ionic radii.

Answer 1

Ionic radii

The ionic radii of isoelectronic species increase with a decrease in the magnitude of nuclear charge. The arrangement of the given species in order of their increasing nuclear charge is as follow:

$\text{Species}{N}^{3–}<O^{2–}<F^{–}<N{a}^{+}<M{g}^{2+}<A{l}^{3+}$

$\text{Nuclear}+7+8+9+11+12+13$
Charge


Therefore, the arrangement of given species in order of their increasing ionic radii is as follow: $A{l}^{3+}<M{g}^{2+}<N{a}^{+}<F^{-}<O^{2-}<N^{3-}$

Final answer:

(a) Number of electrons in all the given species $=10$
Hence these ions are all isoelectronic.

(b) Increasing order of ionic radii is $A{l}^{3+}<M{g}^{2+}<N{a}^{+}<F^{-}<O^{2-}<N^{3-}$