Question

Oxidation number of iodine in IO₃^-, IO₄^-, KI and I₂ respectively is:

• A. -2, -5, -1 ,0
• B. +5, +7, -1, 0
• C. +2, +5, +1, 0
• D. -1, +1, 0, +1

Answer 1

The correct option is B

+5, +7, -1, 0

Explanation for correct answer:

Oxidation Number:

• The total number of electrons that an atom either gains or loses in order to form a chemical bond with another atom is known as the Oxidation Number.
• Generally, the Oxidation number is the charge present on the ions.

We know that charges on Oxygen($\mathrm{O}$) and Potassium($\mathrm{K}$) are -2 and +1 respectively.

Let the oxidation state of Iodine be x.

For ${{\mathrm{IO}}_3}^{-}$

Oxidation state of Iodine + 3 x Oxidation State of Oxygen = -1 (As the total charge on ${{\mathrm{IO}}_3}^{-}$ is -1)

x + 3(-2) = -1

x = +5

Oxidation State of Iodine in ${{\mathrm{IO}}_3}^{-}$ is +5.

For ${{\mathrm{IO}}_4}^{-}$,

Oxidation State of iodine + 4 x Oxidation State of Oxygen = -1 (As the total charge on ${{\mathrm{IO}}_4}^{-}$ is -1)

x + 4(-2) = -1

x = +7

Oxidation State of Iodine in ${{\mathrm{IO}}_4}^{-}$ is +7.

For $\mathrm{KI}$,

Oxidation of Potassium + Oxidation of Iodine = 0 (As the total charge on $\mathrm{KI}$ is 0)

+1 + x = 0

x = -1

Oxidation state of Iodine in $\mathrm{KI}$ is -1.

For ${\mathrm{I}}_2$,

As Iodine is present in nature as ${\mathrm{I}}_2$, so Oxidation state of ${\mathrm{I}}_2$ will be 0.