Question

What is the oxidation number of iodine in ${I_2}$.

Answer 1

Oxidation number is the charge left on the atom when there is cleavage of bond between the atoms. This charge is due to the loose or gain of electrons. When the bond gets broken then one atom acquires positive while the other acquires negative charge. Oxidation number is the charge which an atom gets after cleavage of a bond.

Complete answer:
By the definition of oxidation number we come across that it is the number which an atom gets when it gets separated from the bond formed in between the two atoms. Thus when there is a cleavage of bond between the two atoms, then the atom acquires a charge which is called an oxidation number or sometimes called oxidation state. Iodine molecule${I_2}$ , contains two atoms of iodine. These atoms are joined by covalent bonds. The bond cleavage is of two types:$- 1{\text{ , 0 , + 1}}$
$1.$ Homolytic Cleavage
$2.$ Heterolytic Cleavage
Iodine molecule contains two iodine atoms which means that there is no difference in electronegativity of the two atoms. Thus we can say that the cleavage between the atoms will be a homolytic cleavage. Thus no atom acquires any charge while the bond gets broken. Thus no charge on the atoms which means there is zero charge on atoms. Therefore the oxidation number of the iodine atoms in the iodine molecule is zero. But in the case of $ICl$ there is heterolytic cleavage of bond so iodine will become ${I^ + }$ and chlorine will become $C{l^ - }$. This is because of the electronegativity difference.

Note:
The oxidation number of atoms in a pure molecule is always zero. The cleavage of the bond depends upon the electronegativity difference of the two atoms. Oxidation number of iodine can vary from $- 1$ to $+ 1$. Thus it can have three oxidation states $- 1{\text{ , 0 , + 1}}$. If in a compound oxidation state of iodine comes other than it which means there is mistake.