Q.12:- Consider the following species:
N3–, O2–, F–, Na+, Mg2+ and Al3+
(a) What is common in them?
(b) Arrange them in the order of increasing ionic radii.
N3- has 7+3 = 10 electrons
O2- has 8+2= 10 electrons
F- has 9+1 = 10 electrons
Na+ has 11-1 = 10 electrons
Mg2+ has 12-2 = 10 electrons
Al3+ has 13-3= 10 electrons
(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:
N3– < O2– < 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– < O2– < N3–