Question

Oxidation number of iodine in ${IO_3^-}$, ${IO_4^-}$, $KI$, ${I_2}$ respectively are:
A) -1,-1,0,+1
B) +3,+5,+7,0
C) +5,+7,-1,0
D) -1,-5,-1,0
E) -2,-5,-1,0

Answer 1

Oxidation number is defined as the total number of electrons that an atom either gain or loses in order to form a chemical bond with the atom and form a compound.

Complete step by step answer:
Iodine belongs from the 7A group and it is a non-metal and having oxidation state -1 i.e. it is always in need of 1 electron to complete its octet and come in stable state. For example, if we consider $BrC{l_2}$ in this bromine having oxidation state of +2 because $Cl$ is more electronegative.. For calculating the oxidation number we take all the charges on the right hand side and equal to zero. First, we calculate the oxidation state of $IO_3^ -$.
Oxidation number of oxygen is -2 and this compound has three molecules of oxygen. So the oxidation state is:
Let’s consider the oxidation state of iodine be x,
x + 3( - 2) = - 1
x - 6 = - 1
x = - 1 + 6
x = + 5
Hence, the oxidation number of iodine in $IO_3^ -$ is +5.
Now, calculate oxidation number in $IO_4^ -$ is:
Let’s consider oxidation number of iodine is x,
x + 4( - 2) = - 1
x - 8 = - 1
x = - 1 + 8
x = + 7
Hence, the oxidation state of iodine in $IO_4^ -$ is +7.
Now, calculate oxidation state of iodine in $KI$ is
Let’s consider oxidation state of iodine is x
1 + x = 0
x = - 1
Hence, the oxidation state of iodine in $KI$ is -1.
Now, calculate oxidation state of iodine in ${I_2}$ is
Let’s consider the oxidation state of iodine be x
2x = 0
x = 0
Hence, oxidation state of iodine in ${I_2}$ is 0
Thus, from the above statement we conclude that the oxidation state of $IO_3^ - ,IO_4^ - ,KI\;and\;{I_2}$ is $+ 5, + 7, - 1,0$

Hence, option (C) is the correct answer.

Note: We only calculate the oxidation number when the redox reaction is balanced in nature and then we divide it into two parts, the input and output and finally adding the electrons of equations. Iodine possesses both oxidation states positive and negative due to its electronegativity.