Question

How would you arrange the ions ${N^{ - 3}}$, ${O^{ - 2}}$, $M{g^{2 + }}$, $N{a^ + }$ and ${F^ - }$ in order of increasing ionic radius?

Answer 1

Hint: This question is based on the isoelectronic species as all the ions present are isoelectronic with each other. In case of isoelectronic species lower will be the nuclear charge higher will be the ionic radius.

Complete step by step answer:
The ions are formed by the loss or gain of electrons by the neutral atom to attain the stable configuration. When the atom loses electrons then cation is formed which is a positive charge species. When the atoms gain electrons then anion is formed which is a negative charge species.
The given ions ${N^{ - 3}}$, ${O^{ - 2}}$, $M{g^{2 + }}$, $N{a^ + }$ and ${F^ - }$are isoelectronic species.
The isoelectronic species are defined as those ions which contain the same number of electrons and their electronic configuration is also the same but they differ in the nuclear charge as the number of protons is different as they are derived from different atoms.
The given ions are:
${N^{ - 3}}$-It has 7 protons and 10 electrons.
${O^{ - 2}}$-it has 8 protons and 10 electrons
${F^ - }$-it has 9 protons and 10 electrons
$N{a^ + }$-it has 11 protons and 10 electrons
$M{g^{2 + }}$-it has 12 protons and 10 electrons
For isoelectronic species lower will be the nuclear charge higher will be the ionic radius.
So the increasing order of the ionic radius is
$M{g^{2 + }} < N{a^ + } < {F^ - } < {O^{ - 2}} < {N^{ - 3}}$

Note:
For a neutral atom, the number of protons and the number of electrons are the same. The atomic radius increases as we move from to down in the group and decreases as we move left to right in the period.