Question

Which among the following are intramolecular redox reactions?
A. \[\text{PC}{{\text{l}}_{5}}\text{ }\to \text{ PC}{{\text{l}}_{3}}\text{ + C}{{\text{l}}_{2}}\]
B. \[\text{CHO - CHO + O}{{\text{H}}^{-}}\text{ }\to \text{ CO}{{\text{O}}^{-}}-\text{C}{{\text{H}}_{2}}\text{OH}\]
C. \[2\text{KCl}{{\text{O}}_{3}}\text{ }\to \text{ 2KCl + 3}{{\text{O}}_{2}}\]
D. \[\text{N}{{\text{H}}_{5}}\text{N}{{\text{O}}_{2}}\text{ }\to \text{ }{{\text{N}}_{2}}\text{ + 2}{{\text{H}}_{2}}\text{O}\]

Answer 1

Hint: For this problem, we should know about the intramolecular redox reactions and how to identify them. Then by studying all the reactions we can choose the correct options in which both oxidation and reduction processes take place.

Complete Step-by-Step answer:
- In the given question, we have to identify the correct example of intramolecular redox reaction among the given options.
- As we know that the intramolecular redox reaction is a type of reaction in which the oxidation and reduction reaction takes place within the same molecule.
- So, among the given options all are the correct answer. As in reaction first i.e.
\[\text{PC}{{\text{l}}_{5}}\text{ }\to \text{ PC}{{\text{l}}_{3}}\text{ + C}{{\text{l}}_{2}}\]
- Phosphorus undergoes reduction because a decrease in the oxidation state takes place from +5 to +3.
In \[\text{PC}{{\text{l}}_{5}}\], \[\text{x + (5 }\times \text{ -1) = 0}\] ; \[\text{x = +5}\]
In \[\text{PC}{{\text{l}}_{3}}\], $\text{x + (3 }\times \text{ -1) = 0}$; $\text{x = +3}$
- Whereas chlorine undergoes oxidation because of the increase in the oxidation state from -1 to 0.
In \[\text{PC}{{\text{l}}_{5}}\], $5\,\text{+ 5x = 0}$; $\text{x = -1}$
In \[\text{C}{{\text{l}}_{2}}\], the oxidation state is zero because it is a single atom.
- So, this reaction is an example of an intramolecular redox reaction.
- Now, in the third reaction also both reduction and oxidation take place i.e.
\[2\text{KCl}{{\text{O}}_{3}}\text{ }\to \text{ 2KCl + 3}{{\text{O}}_{2}}\]
- oxygen undergoes oxidation because of the increase in the oxidation state from -2 to 0.
In \[\text{KCl}{{\text{O}}_{3}}\], the oxidation state of chlorine is -2 because when oxygen is present with the less electronegative atom it shows -2 oxidation state.
In \[{{\text{O}}_{2}}\], the oxidation state is zero because it is a single atom.
- Whereas chlorine undergoes reduction because of the decrease in the oxidation state that takes place from +5 to -1.
In \[\text{KCl}{{\text{O}}_{3}}\], $1\text{ + x + (3 }\times \text{ -2) = 0}$; $\text{x = +5}$
In \[\text{KCl}\], $1\text{ + x = 0}$; $\text{x = -1}$
- So, this reaction is an example of the intramolecular redox reaction.
- Now, in forth reaction, the intramolecular redox reaction takes place because the oxidation, as well as reduction of the phosphorus, take place.
\[\text{N}{{\text{H}}_{4}}\text{N}{{\text{O}}_{2}}\text{ }\to \text{ }{{\text{N}}_{2}}\text{ + 2}{{\text{H}}_{2}}\text{O}\]
In \[\text{N}{{\text{H}}_{5}}\text{N}{{\text{O}}_{2}}\], the oxidation state of first nitrogen is \[{}^{1}\text{N}{{\text{H}}_{4}}\text{N}{{\text{O}}_{2}}\ \text{= x + 4 }\times \text{ 1 + -1 = 0}\]
\[\text{x = -3}\].
- Now, the oxidation state of second nitrogen is \[\text{N}{{\text{H}}_{4}}{}^{2}\text{N}{{\text{O}}_{2}}\text{, 0 + x + (2 }\times \text{ -2) = 0}\]; \[\text{x = +4}\]
Whereas the oxidation state of nitrogen is zero.
- So, we can say the nitrogen undergoes reduction as well as oxidation.
- In the second reaction, it is an example of Cannizzaro reaction in which one aldehyde is oxidised to carboxylic acid whereas another one is reduced to an alcohol.

Therefore, all the given reactions show intramolecular redox reactions.

Note: In intermolecular redox reaction the oxidation and reduction reaction take place between two or more molecules. The second reaction is also an example of a disproportionation reaction.