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

## Chapter 8 - Redox Reactions Exercise 280

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 281

## Chapter 8 - Redox Reactions Exercise 282

## Chapter 8 - Redox Reactions Exercise 283

### Kindly Sign up for a personalised experience

- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions

#### Sign Up

#### Verify mobile number

Enter the OTP sent to your number

Change