Question

Correct radii order is
A. ${{P}^{3-}}>S{{e}^{2-}}>{{I}^{-}}$
B. ${{P}^{3-}}>{{I}^{-}}>S{{e}^{2-}}$
C. ${{I}^{-}}>{{P}^{3-}}>S{{e}^{2-}}$
D. ${{I}^{-}}>S{{e}^{2-}}>{{P}^{3-}}$

Answer 1

Hint: When an atom loses or gains an electron then it forms ions. When it loses an electron it forms a positively charged ion which is named as cation and on the other hand if an atom gains an electron then it forms a negatively charged ion known as anion. The atoms given in the question are anions i.e. negatively charged ions.

Complete step by step answer:
- Ionic radius can be defined as the distance between the nucleus of an ion to the outermost shell of that ion. In periodic table while moving down the group ionic radii increase due to increase in number of shells and on moving across the period atomic radius first decreases then increases sharply and then again decreases.
- In the given question the ionic radii follows the pattern ${{P}^{3-}}>S{{e}^{2-}}>{{I}^{-}}$ this order can be explained on the basis of moving down the group as $P$ represented phosphorus is group 3rd element, $Se$ which represents selenium is from group 4 and $I$ representing iodine is of group 5 and we know that if we move down the group then ionic radii increases as it is distance between the nucleus and outermost shell and in this case the number of shells increases so as a result ionic radii also goes on increasing.
The correct answer is option “A” .

Note: Ionic radius of cation is always smaller than ionic radius of anion this can be explained as cation will have a greater positive charge, so it will attract the electrons in the outermost shell with greater force and as a result size becomes smaller as compare to anion.