Question

The halogen with the smallest covalent radius is:
(A) I
(B) Cl
(C) Br
(D) F

Answer 1

Covalent radius of an atom is equal to the half of the length of a covalent bond. The sum of the covalent radii of two atoms forming a covalent bond is equal to the length of their covalent bond.

Complete step by step solution:
Here, we need to find the halogen atom given in the option that has the smallest covalent radius. Let’s know about the covalent radius of the atom.
- Covalent radius of an atom is equal to the half of the length of a covalent bond. That means the sum of the covalent radii of two atoms forming a covalent bond is equal to the length of their covalent bond. Now, let’s compare the covalent radius of given halogen atoms.
- I (Iodine) is placed in the fifth period of the periodic table of elements. We know that the atomic size increases as we go down in the periodic table. This is due to the increase in the number of shells down the group. Same thing is for covalent radius. It also increases as we go down in the periodic table. So, amongst given options, iodine should have the highest covalent radius.
- Br (Bromine) is in the fourth period of the periodic table. So, its covalent radius is smaller than iodine but bigger than chlorine and fluorine.
- Cl (chlorine) is placed in the third period of the periodic table. Its covalent radius is smaller than Br and I but bigger than fluorine.
- F (Fluorine) is placed in the second period in the periodic table. It is the halogen with the smallest covalent radius.

Therefore, the correct answer is (D).

Note: We can arrange the halogens in the order of decreasing covalent radii as:
I > Br > Cl > F
Note that the covalent radius is smaller than the ionic radius in the case of halogens.