# NCERT Solutions for Class 11-science Chemistry Chapter 8 - Redox Reactions

## Chapter 8 - Redox Reactions Exercise 272

a. NaH_{2}PO_{4}

Let the oxidation number of P be *x*.

We know that,

Oxidation number of Na = +1

Oxidation number of H = +1

Oxidation number of O = –2

Then, we have

1(+1) +2(+1)+1(x)+4(-2) =0

1+2+x-8 = 0

x = +5

Hence, the oxidation number of P is +5.

b. NaHSO_{4}

Hence, the oxidation number of S is + 6.

c. H_{4}P_{2}O_{7}

Hence, the oxidation number of P is + 5.

d. K_{2}MnO_{4}

Hence, the oxidation number of Mn is + 6.

e. CaO_{2}

Hence, the oxidation number of O is – 1.

f. NaBH_{4}

Hence, the oxidation number of B is + 3.

g. H_{2}S_{2}O_{7}

Hence, the oxidation number of S is + 6.

h. KAl(SO_{4})_{2}.12H_{2}O

Or,

We can ignore the water molecule as it is a neutral molecule. Then, the sum of the oxidation numbers of all atoms of the water molecule may be taken as zero. Therefore, after ignoring the water molecule, we have

Hence, the oxidation number of S is + 6.

a. KI_{3}

_{3}, the oxidation number (O.N.) of K is +1. Hence, the average oxidation number of I is . However, O.N. cannot be fractional. Therefore, we will have to consider the structure of KI

_{3}to find the oxidation states.

_{3}molecule, an atom of iodine forms a coordinate covalent bond with an iodine molecule.

_{3}molecule, the O.N. of the two I atoms forming the I

_{2}molecule is 0, whereas the O.N. of the I atom forming the coordinate bond is –1.

b. H_{2}S_{4}O_{6}

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. Fe_{3}O_{4}

On taking the O.N. of O as –2, the O.N. of Fe is found to be . 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.** **CH_{3}CH_{2}OH

Hence, the O.N. of C is –2.

e. CH_{3}COOH

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 CH_{3}COOH.

## Chapter 8 - Redox Reactions Exercise 273

## Chapter 8 - Redox Reactions Exercise 274

## Chapter 8 - Redox Reactions Exercise 275

### Other Chapters for CBSE Class 11-science Chemistry

Chapter 1- Some Basic Concepts of Chemistry Chapter 2- Structure Of The Atom Chapter 3- Classification of Elements and Periodicity in Properties Chapter 4- Chemical Bonding and Molecular Structure Chapter 5- States of Matter Chapter 6- Thermodynamics Chapter 7- Equilibrium Chapter 9- Hydrogen Chapter 10- The s-Block Elements Chapter 11- The p-Block Elements Chapter 12- Organic Chemistry: Some Basic Principles and Techniques Chapter 13- Hydrocarbons Chapter 14- Environmental Chemistry### NCERT Solutions for CBSE Class 11-science Subjects

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