Question

arrange the following in increasing order of ionic radii : n3-, o2-, na+ , al3+, mg2+

Answer 1

Since all these ions have 10 electrons in their shell therefore these are isoelectronic speicies

The more + the charge, the smaller the ionic radius. Remember that - means adding electrons. These electrons go in the outermost shells. Also, when an atom loses electrons, it clings ever more tightly to the ones it has left, further reducing the ionic radius. therefore the order of ionic radii will be →

AL3+ Mg2+ Na+ F- O2- N3- (increasing order)

Thanks

Answer 2

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

Explanation:

Isoelectronic specie are defined as the molecules which have the same number of electrons.

Atomic number of nitrogen is 7 and thus has 7 electrons.$N^{3-}$ is formed by gain of 3  electrons and thus has 10 electrons.

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

Atomic number of sodium is 11 and thus has 11 electrons. $Na^{+}$  is formed by loss of 1 electron and thus has 10 electrons.

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

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

As the number of electrons are same , the more is the nuclear charge i.e. the number of protons , more will the force of attraction towards the nuclei and thus smaller will be the radii and vice versa.

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

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