Question

From the above mentioned periodic table (abbreviated), the element with the smallest ionic radius is:

(A) A
(B) B
(C) C
(D) D
(E) E

Answer 1

Hint:Ions are the charged species : there are two type of ions ie. Cation and anion. Ionic radius is measured as the distance between cations and anions in an ionic crystal. Ionic radii show different trends along the period and down the group.

Complete step by step answer:
The trends shown by ionic radius are :-
1. Down the group :-
When we move down the group, new shells are added in the atoms and the number of electrons increases. Due to addition of shells, the ionic radius of atoms increases.
2. Across a period :-
As we move from left to right in a group, the ionic radius first decreases then increases and then again decreases. For metals the ionic radius decreases as cations are formed by them. Metals lose their electron (S) to form cation due to which, nuclear change per electron increases and electrons get tightly held within the atom. That is why it is said that cations are smaller than their parent atom.
For non-metals, ionic radius increases as they form anions nonmetals gain electron (S) to form anions due to which the nuclear charge per electron decreases and electrons become less tightly held within the atom. Therefore, it is said that anions are larger in size as compared to its parent atom.
In the given question, the elements A and C are present below the D. and in turn B and E are present below A and C. Due to the position down in the group, their ionic radius goes on increasing.
Whereas, if we see across the period the atoms are arranged as B, then A, then C, then D and E at last.
E is a noble gas as it falls in the group of neon (a noble gas).
Therefore, we can say that the element D has the smallest ionic radius because it is present in the top period and second last group.
Hence, the correct option is (D) .


Note:
In anions, the ionic radii is larger than its parent atom. This is because the net repulsion of electrons will outweigh the nuclear charge and the ion will become large in size.