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Academic session 2026-27, Chapter 7 in the current curriculum.
Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter Redox Reactions. This page offers a step-by-step solution to the specific question from Exercise 1, Question 2:
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
. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.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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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 2: What are the oxidation number of the underlined elements in each of the following and how do you rat....
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Thanks for your NCERT solution