Chemistry Class 11 NCERT Solutions: Chapter 8 Redox Reactions Part 2

Chemistry Class 11 NCERT Solutions Chapter 8 Redox Reactions Part 2

Q: 2. What are the oxidation numbers of the underlined elements in each of the following and how do you rationalize your results?

(A) Equation

(B) Equation

(C) Equation

(D) Equation

(E) Equation

Answer:

(A) Equation

In Equation the oxidation number (O.N.) of K is + 1. Hence, the average oxidation number of I is Equation . However, O.N. cannot be fractional. Therefore, we will have to consider the structure of Equation to find the oxidation states

In a Equation molecule, an atom of iodine forms a coordinate covalent bond with an iodine

Molecule.

Equation

Hence, in a Equation molecule, the O.N. of the two I atoms forming the Equation molecule is 0, whereas the O.N. of the I atom forming the coordinate bond is –1.

(B) Equation

Equation

Now, Equation

Equation

Equation

Equation

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) Equation

On taking the O.N. of O as –2, the O.N. of Fe is found to be Equation . 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.

Equation

(D) Equation

Equation

Equation

Equation

Equation

Hence, the O.N. of C is -2.

(E) Equation

Equation

Equation

Equation

Equation

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 Equation and –2 in Equation .

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