Question

Arrange the following species in the increasing order of their ionic radii, giving reasons.
$M{g^{2 + }},{O^{2 - }},A{l^{3 + }},{N^{3 - }},{F^ - },N{a^ + }$

Answer 1

To solve the question we need to understand the concept of ionic radii and the factors affecting it. In a neutral atom, Atomic and Ionic radii are the distance between the nucleus and the outermost electron. However, when the atom loses or gains electrons it forms Ions i.e. cations and anions respectively. Anion is further away from the nucleus so is large while cation is closer to the nucleus hence smaller than the neutral atom. The correct sequence of the size of cation, anion, and neutral atom is -Cation
Complete answer:
In the question we are given certain species, let us first count the number of electrons they have-
$M{g^{2 + }}$ - Here Magnesium has an atomic number (Number of protons) of 12 and a +2 charge means 2 electrons less than the neutral atom so it has 10 electrons.
${O^{2 - }}$ - Here Oxygen has an atomic number 8 and a -2 charge means 2 electrons more than the neutral atom so it has 10 electrons.
$A{l^{3 + }}$ - Here Aluminium has an atomic number 13 and a +3 charge means 3 electrons less than the neutral atom so it has 10 electrons.
${N^{3 - }}$ - Here Nitrogen has an atomic number 7 and a -3 charge means 3 electrons more than the neutral atom so it has 10 electrons.
${F^ - }$ - Here Fluorine has an atomic number 9 and a -1 charge means 1 electron more than the neutral atom so it has 10 electrons.
$N{a^ + }$ - Here Sodium has an atomic number 11 and a +1 charge means 1 electron less than the neutral atom so it has 10 electrons.
So we see that the species are isoelectronic (having the same number of electrons) so the deciding factor is the number of protons or positive charge (effective nuclear charge). The more the atomic number, the more the number of protons, the attraction is more and hence smaller the size. In the case of isoelectronic species, we need not consider the overall charge on the ion and only look at the positive charge in the nucleus. Here Aluminium ions have the most positive charge and the Nitrogen ion has the least positive charge.
The correct sequence will be thus as follows-
$A{l^{3 + }} < M{g^{2 + }} < N{a^ + } < {F^ - } < {O^{2 - }} < {N^{3 - }}$ .

Note:
The Ionic and atomic radii increase from top to bottom in a group because as we move down the group the number of shells increases and the interelectronic repulsion also increases. The ionic and atomic radii decrease from left to right in a period because from left to right the effective nuclear charge increases and thus the attractive forces overcome the repulsive forces between electrons thus reducing the size.