Consider the following species:

N^3–, O^2–, F^–, Na⁺, Mg²⁺ and Al³⁺

(a) What is common in them?

(b) Arrange them in the order of increasing ionic radii.

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Solution

(a) Each of the given species (ions) has the same number of electrons (10 electrons). Hence, the given species are isoelectronic.

(b) The ionic radii of isoelectronic species increases with a decrease in the magnitudes of nuclear charge.

The arrangement of the given species in order of their increasing nuclear charge is as follows:

N^3– < O^2– < F^– < Na⁺ < Mg²⁺ < Al³⁺

Nuclear charge = +7 +8 +9 +11 +12 +13

Therefore, the arrangement of the given species in order of their increasing ionic radii is as follows:

Al³⁺ < Mg²⁺ < Na⁺ < F^– < O^2– < N^3–

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Similar questions

Consider the following species:

N^3–, O^2–, F^–, Na⁺, Mg²⁺ and Al³⁺

(a) What is common in them?

(b) Arrange them in the order of increasing ionic radii.

N^{3 -}

O^{2 -}

F^{-}

{N a}^{+}

{M g}^{2 +}

{A l}^{3 +}

N^{3 -}, O^{2 -}, M g^{2 +}

F^{-}

{M g}^{2 +}, O^{2 -} {A l}^{3 +}; N^{3 -}, F^{-}, {N a}^{+}

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