Question

How many single bonds can iodine form?

Answer 1

Determine the chemical representation of the iodine atom. Using the periodic table calculate the total number of valence electrons present in iodine. Using the octet rule determines the valency of the iodine atom.

Complete step-by-step answer: The atomic symbol of iodine is ${\text{I}}$. By referring to periodic tables we can say that element chlorine (${\text{I}}$) is a halogen present in the VIIA group and the 5th period of the periodic table. To determine the number of single bonds can form iodine we need to determine the number valency of an iodine atom .Valence electrons are the outer shell electrons that participate in the formation of the chemical bond.
The atomic number of ${\text{I}}$ = 53
The electronic configuration of iodine is as follows:
$[{\text{Kr]4}}{{\text{d}}^{{\text{10}}}}{\text{5}}{{\text{s}}^{\text{2}}}{\text{5}}{{\text{p}}^{\text{5}}}$
From the electronic configuration of iodine, we can say that ${\text{5}}{{\text{s}}^{\text{2}}}{\text{5}}{{\text{p}}^{\text{5}}}$ are outer shell electrons of iodine so there are 7 valence electrons present in iodine. The octet rule states that atoms always tend to lose, gain or share electrons to achieve the electronic configuration of the nearest inert gas. From the electronic configuration of the iodine atom, we can say that the valency of the iodine atom is 1 as it tends to gain one electron to complete its octet. Valency is the combining capacity of the atom. It gives us information about the number of chemical bonds can atom form. As the valency of iodine is 1 we can say that iodine can form one single bond with another atom. Thus, iodine can form only one single bond.

Note: Atoms having more than 4 valence electrons always tend to gain electrons to complete their octet. Atoms having less than 4 valence electrons always tend to lose their valence electrons to complete their octet. Metals tend to lose their valence electrons while nonmetals tend to gain or share their valence electrons.