Question

The oxidation state of iodine in ${{H}_{4}}I{{O}_{6}}^{-}$ is:
a) +7
b) -1
c) +5
d) +1

Answer 1

To calculate oxidation state it is important to understand how to assign oxidation number to the elements. If we know the oxidation state of other elements in the given substance then we can easily calculate the oxidation state of the remaining element.

Complete Solution:
There are certain rules to assign oxidation number to the elements which are mentioned below-
Rule 1- The oxidation of an element in uncombined or Free State is zero.
Rule 2- for the monatomic (only one atom) ion the oxidation state is the same as the charge of the ion.
Rule 3- In a polyatomic (more than one) ion the sum of all the oxidation states is equal to the charge on the ions.
Rule 4- The oxidation number of alkali metal and alkaline earth metals are +1 and +2 respectively.
Rule 5- The oxidation number of oxygen in a compound is usually -2.
Rule 6- The oxidation state of hydrogen is usually +1.
Rule 7- The oxidation state of halogens is usually -1.
Calculation of oxidation state of iodine ${{H}_{4}}I{{O}_{6}}^{-}$ is
Oxidation state of H= +1
Oxidation state of O = -2
Let Oxidation state of I = x

& 4(1)+x+6(-2)=-1 \\\ & 4+x-12=-1 \\\ & x-8=-1 \\\ & x=+7 \\\ \end{aligned}$$ Hence the correct option is +7 **So, the correct answer is “Option A”.** **Note:** The oxidation state of oxygen is not always -2 in case of peroxides the oxidation state of oxygen is not -2 similarly when oxygen is bonded to fluorine the oxidation state is +1. \- Same for the case of hydrogen when hydrogen is bonded to binary metal halide its oxidation state is -1 and when the halogens are bonded to oxygen or fluorine their oxidation state is not -1.