# Chemistry Class 11 NCERT Solutions: Chapter 8 Redox Reactions Part 6 (For CBSE, ICSE, IAS, NET, NRA 2022)

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Q: 7. Suggest a list of the substances where carbon can exhibit oxidation states from and nitrogen from .

The substances where carbon can exhibit oxidation states from are listed in the following table.

 Substance O. N. of Carbon 0 + 1 -1 + 2 -2 + 3 -3 + 4 -4

The substances where nitrogen can exhibit oxidation states from – 3 to + 5 are listed in the following table.

 Substance O. N. of Nitrogen 0 + 1 -1 + 2 -2 + 3 -3 + 4 + 5

Q: 8. while sulphur dioxide and hydrogen peroxide can act as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why?

In sulphur dioxide, the oxidation number (O. N.) of S is + 4 and the range of the O. N. that S can have is from + 6 to – 2.

Therefore, can act as an oxidising as well as a reducing agent.

In hydrogen peroxide , the O. N. of O is – 1 and the range of the O. N. that O can have is from 0 to – 2. O can sometimes also attain the oxidation numbers + 1 and + 2. Hence, can act as an oxidising as well as a reducing agent.

In ozone () , the O. N. of O is zero and the range of the O. N. that O can have is from 0 to – 2. Therefore, the O. N. of O can only decrease in this case. Hence, acts only as an oxidant.

In nitric acid, the O. N. of N is + 5 and the range of the O. N. that N can have is from + 5 to – 3. Therefore, the O. N. of N can only decrease in this case. Hence, acts only as an oxidant.

Q: 9. Consider the reactions:

(A)

(B)

Why it is more appropriate to write these reactions as:

(A)

(B)

Also, suggest a technique to investigate the path of the above (a) and (b) redox reactions.

(A) The process of photosynthesis involves two steps.

Step 1:

Decomposes to give and .

Step 2:

Now, the net reaction of the process is given as:

It is more appropriate to write the reaction as given above because water molecules are also produced in the process of photosynthesis.

The path of this reaction can be investigated by using radioactive in place of .

(B) is produced from each of the two reactants and . For this reason, is written twice.

The given reaction involves two steps. First, decomposes to form and O. In the second step, reacts with the O produced in the first step, thereby producing and .

The path of this reaction can be investigated by using

Q: 10. The compound is an unstable compound. However, if formed, the compound acts as a very strong oxidizing agent. Why?