Question

Question 3.12 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.

Class XI Classification of Elements and Periodicity in Properties Page 93

Answer 1

(a) N³⁻ ( 10 electrons ) , O²⁻ (10 electrons ),
F⁻ ( 10 electrons ), Na⁺ (10 electrons ),
Mg²⁺(10 electrons) , Al³⁺(10electrons)

we can see that all the given species have same number of electrons ( 10 electrons )
so, all are isoelectronic species .

(b) we know, The ionic radii of isoelectronic species decreases with increase in atomic number because magnitude of nuclear charge increases with increase atomic number .

increasing order of radii ,
Al³⁺(Z = 13) < Mg²⁺( Z = 12 ) < Na⁺ ( Z = 11) < F⁻ (Z = 9) < O²⁻(Z = 8) < N³⁻ (Z = 7)

Answer 2

Explanation:

(a)    Atomic number of nitrogen is 7 and $N^{3-}$ has 10 electrons.

Atomic number of oxygen is 8 and $O^{2-}$ has 10 electrons.

Atomic number of fluorine is 9 and $F^{-}$ has 10 electrons.

Atomic number of sodium is 11 and $Na^{+}$ has 10 electrons.

Atomic number of magnesium is 12 and $Mg^{2+}$ has 10 electrons.

Atomic number of aluminium is 13 and $Al^{3+}$ has 10 electrons.

Therefore, in all the given ions it is common that all of them have 10 electrons.

(b)    It is known that more is the positive charge on an ion smaller will be its ionic radii. Whereas more is the negative charge on an ion more will be its ionic radii.

Thus, order of increasing ionic radii for the given ions is as follows.

 $Al^{3+}$ < $Mg^{2+}$ < $Na^{+}$ < $F^{-}$ < $O^{2-}$ < $N^{3-}$