Question

Oxidation number of iodine varies from:
A. -1 to +1
B. -1 to +7
C. +3 to +5
D. -1 to +5

Answer 1

Hint: Oxidation number is the number of electrons that the atom of an element loses or gains during chemical bond formation. The oxidation number can be zero, negative, or positive.

Complete step by step answer:
The atomic symbol of iodine is ${\text{I}}$ . It is a group 7 element, belonging to the halogen group. The atomic number of iodine is 53. As it is the seventh group it has 7 valence electrons.

The electronic configuration of iodine is as follows:

${\text{[Kr] 4}}{{\text{d}}^{{\text{10}}}}{\text{5}}{{\text{s}}^{\text{2}}}{\text{5}}{{\text{p}}^{\text{5}}}$

Outermost ${\text{5}}{{\text{s}}^{\text{2}}}{\text{5}}{{\text{p}}^{\text{5}}}$ electrons are known as valence electrons.
As there are 7 valence electrons in iodine it tends to either gain 1 electron or lose 7 electrons to complete its octet.
When there is a loss of electron oxidation number is positive and when there is a gain of electron oxidation number is negative.
To complete the octet iodine can gain only one electron. Thus, the possible negative value of the oxidation number of iodine is -1.
To complete the octet iodine can also lose its 7 valence electrons so possible positive values of oxidation number of iodine are +1,+2,+3,+4,+5,+6 and +7.
Thus, the oxidation number of iodine varies from -1 to +7.

So, the correct option is (B) -1 to +7.

Note:
Oxidation number is never the actual charge over the atom, except in ionic compounds. It is the only apparent charge which represents the positive or negative character of the atom. Oxidation is the loss of electrons while reduction is the gain of electrons. During oxidation, the oxidation number of an atom increases while during reduction-oxidation number of an atom decreases.