Question

The oxidation state of iodine in ${H_4}I{O_6}^ -$ is ?
a.) +7
b.) -1
c.) +5
d.) +1

Answer 1

Hint : The oxidation state of iodine is equal to the number of its valence electrons. It can be calculated by adding the known values and putting it equal to (-1).

Complete step by step answer :
We will find oxidation state step by step as –
Initially, we will suppose the oxidation state of I in ${H_4}I{O_6}^ -$ be x.
Now, we know that the overall molecule has a charge of -1. So, the sum of oxidation states of all elements will be equal to -1.
Hydrogen will be +1 *4=4
Oxygen will be (-2)*6=(-12)
Thus, the sum is 4+x-12=(-1)
x-8=(-1)
x=+7

Therefore, the oxidation state of Iodine is +7 and option a.) is the correct answer.

Note : If the molecule would have been neutral; we would have put it equal to zero.
The ${H_4}I{O_6}^ -$ is called Orthoperiodate. It is a monovalent anion obtained after deprotonation. It is a conjugate base of orthoperiodic acid. The ${H_4}I{O_6}^ -$ has a pKa value of 8.31. The orthoperiodic acid has monoclinic structure. Like all other periodic acids, it is used in determining the structure of carbohydrates.