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