Class 11-science NCERT Solutions Chemistry Chapter 7: Redox Reactions

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Redox Reactions Exercise 252

Solution 1

a. NaH₂PO₄

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

$begin mathsize 14px style stack rightwards double arrow Na with plus 1 on top stack straight H subscript 2 with plus 1 on top straight P with x on top straight O with negative 2 on top subscript 4 end style$

Then, we have

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

$rightwards double arrow$ 1+2+x-8 = 0

$begin mathsize 14px style rightwards double arrow end style$ x = +5

Hence, the oxidation number of P is +5.

b. NaHSO₄

Hence, the oxidation number of S is + 6.

c. H₄P₂O₇

Hence, the oxidation number of P is + 5.

d. K₂MnO₄

Hence, the oxidation number of Mn is + 6.

e. CaO₂

Hence, the oxidation number of O is – 1.

f. NaBH₄

Hence, the oxidation number of B is + 3.

g. H₂S₂O₇

Hence, the oxidation number of S is + 6.

h. KAl(SO₄)₂.12H₂O

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.

Solution 2

a. KI₃

In KI₃, the oxidation number (O.N.) of K is +1. Hence, the average oxidation number of I is

begin mathsize 12px style negative 1 third end style

. 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.

begin mathsize 12px style stack straight K to the power of plus with plus 1 on top open square brackets straight I with 0 on top minus straight I with 0 on top leftwards arrow straight I with negative 1 on top close square brackets to the power of minus end style

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

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

On taking the O.N. of O as –2, the O.N. of Fe is found to be $begin mathsize 12px style plus 2 2 over 3 end style$ . 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

Hence, the O.N. of C is –2.

e. CH₃COOH

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.