Question

The correct order for decreasing atomic radius of the five atoms H, Br, I, He, S is:
A.I > Br > S > He > H
B.I > S > Br > H > He
C.I > S > Br >He > H
D.Br > I > S > He > H

Answer 1

Hint: To solve these types of questions you should know about the properties of groups and periodic table periods, e.g. where the size, radius, etc. decreases or increases from.

Complete answer:
As we know, that an atom gets larger as it’s the number of shells increases .
Therefore the radius of atoms increases as you go down a certain group in the periodic table of elements.
As per for the periods, when we move left to right in a particular period the size of an atom decreases.
So, according to this concept, I and Br belong to the halogen family of the 4th and 5th period so they are larger in atomic size, S belongs to the 3rd period and group 16, He is a noble gas with two electrons and similar in size but H with one electron is of least atomic size.
Therefore the decreasing order of atomic radius is I >Br >S >He >H
Hence, Option A is correct which is I >Br >S >He >H.

NOTE: Option B, C, D is wrong because the elements of 4th and 5th period are larger in size of atomic radius as compared to period 3rd and helium which is a noble gas and hydrogen with only one electron in its shell, as the force of attraction increases between nuclei and electrons the size of the atoms decreases. The effect decreases when one travels further to the right in a time due to repulsions from the electron-electron which otherwise would cause the size of the atom to increase.