Chemistry Class 11 NCERT Solutions: Chapter 8 Redox Reactions Part 2 (For CBSE, ICSE, IAS, NET, NRA 2022)

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

(A)

(B)

(C)

(D)

(E)

Answer:

(A)

In 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 to find the oxidation states

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

Molecule.

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

(B)

Now,

However, O. N. cannot be fractional. Hence, S must be present in different oxidation states in the molecule.

Q_2_B_The Oxidation Number of Each of S Atom Linked with Eac …

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)

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)

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

(E)

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 and – 2 in .

Q_2_E_Structure of C Exhibits the Oxidation States of + 2 and …

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