What are the oxidation number of the und | Class 11 Chemistry Chapter Redox Reactions, Redox Reactions NCERT Solutions

Current NCERT chapter

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.

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

Answer:

(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. 

 


Study Tips for Answering NCERT Questions:

NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:

  • Read the question carefully and focus on the core concept being asked.
  • Reference examples and data from the chapter when answering questions about Redox Reactions.
  • Review previous year question papers to get an idea of how such questions may be framed in exams.
  • Practice answering questions within the time limit to improve your speed and accuracy.
  • Discuss your answers with your teachers or peers to get feedback and improve your understanding.

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Comments

  • Deepak rao
  • May 19, 2019

Thanks for your NCERT solution


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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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