Class 11 Chemistry - Chapter Redox Reactions NCERT Solutions | What are the oxidation number of the und
What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results ?
(a) KI3
(b) H2S4O6
(c) Fe3O4
(d) CH3CH2OH
(e) CH3COOH
(a) KI3
Let assume oxidation number of l is x.
In KI3, the oxidation number (O.N.) of K is +1.
1(+1) + 3(x) = 0
⇒ +1 +3x = 0
⇒ 3x = -1
⇒ x = -1/3
Hence, the average oxidation number of I is - 1/3
However, O.N. cannot be fractional. Therefore, we will have to consider the structure of KI3 to find the oxidation states. In a KI3 molecule, an atom of iodine forms a coordinate covalent bond with an iodine molecule.
Hence, in a KI3 molecule, the O.N. of the two I atoms forming the I2 molecule is 0, whereas the O.N. of the I atom forming the coordinate bond is –1.
(b) H2S4O6
Let assume oxidation number of S is x.
The oxidation number (O.N.) of H is +1.
The oxidation number (O.N.) of O is -2.
2(+1) + 4(x) + 6(-2) = 0
⇒ 2 + 4x - 12 = 0
⇒ 4x -10 = 0
⇒ 4x = +10
⇒ x = +10/4
However, O.N. cannot be fractional. Hence, S must be present in different oxidation states in the molecule.
The O.N. of two of the four S atoms is +5 and the O.N. of the other two S atoms is 0.
(c) Fe3O4
Let assume oxidation number of Fe is x.
The oxidation number (O.N.) of O is -2.
3(x) + 4(-2) = 0
⇒ 3x - 8 = 0
⇒ 3x = 8
⇒ x = 8/3
However, O.N. cannot be fractional.
Here, one of the three Fe atoms exhibits the O.N. of +2 and the other two Fe atoms exhibit the O.N. of +3.
(d) CH3CH2OH
Let assume oxidation number of C is x.
The oxidation number (O.N.) of O is -2.
The oxidation number (O.N.) of H is +1.
x + 3(+1) + x + 2(+1) + 1(-2) + 1(+1) = 0
⇒ x +3 + x +2 - 2 + 1 = 0
⇒ 2x + 4 = 0
⇒ 2x = -4
⇒ x = -2
Hence, the oxidation number of C is -2.
(e) CH3COOH
Let assume oxidation number of C is x.
The oxidation number (O.N.) of O is -2.
The oxidation number (O.N.) of H is +1.
x + 3(+1) + x + (-2) + (-2) + 1(+1) = 0
⇒ 2x + 3 - 2 - 2 + 1 = 0
⇒ 2x + 0 = 0
⇒ x = 0
However, 0 is average O.N. of C.
The two carbon atoms present in this molecule are present in different environments. Hence, they cannot have the same oxidation number. Thus, C exhibits the oxidation states of +2 and –2 in CH3COOH.
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