Question

Select the an intramolecular redox reactions among the following

$$A.{\text{ }}2KCl{O_3} \to 2KCl + 3{O_2} \\\ B.{\text{ }}{(N{H_4})_2}C{r_2}{O_7} \to {N_2} + C{r_2}{O_3} + 4{H_2}O \\\ C.{\text{ }}C{l_2} \to Cl + Cl{O_3}^ - \\\ D.{\text{ }}N{H_4}N{O_2} \to {N_2} + 2{H_2}0 \\\$$

Answer 1

Hint- We must remember that in an intramolecular redox reaction, the molecule of a single substance undergoes oxidation-reduction by the decomposition process. Therefore, we can find the intramolecular redox reaction based on the oxidation and reductions of reactants and products in the given reactions.

Complete step by step answer:
(i) In this reaction, we can understand that the potassium gains six electrons to get reduced from +5 states to -1 state and oxygen loses two electrons to get oxidized from -2 states. Therefore $KCl{O_3}$ undergoes oxidation-reduction by the decomposition process. Hence, this is an intramolecular redox reaction.
${\text{ }}( + 5){\text{ }}( - 2){\text{ }}( - 1){\text{ }}(0) \\\ 2KCl{O_3} \to 2KCl + 3{O_2} \\\$
(ii) In the second reaction, we can easily understand that the reduction process only occurs. Therefore, nitrogen (N) gained three electrons and chromium (Cr) gained three electrons in${(N{H_4})_2}C{r_2}{O_7}$. Hence, the nitrogen and chromium gets reduced from +3 and +6 oxidation states to 0 and +3 states respectively. So we can conclude that this is not an intramolecular redox reaction.

$${\text{ }}( + 3){\text{ }}( + 6){\text{ }}(0){\text{ }}( + 3) \\\ {(N{H_4})_2}C{r_2}{O_7} \to {N_2} + C{r_2}{O_3} + 4{H_2}O \\\$$

(iii) The reaction of C is an intramolecular reaction. In this reaction, we can see that the chlorine (Cl) gets oxidations and reductions simultaneously. Therefore the chlorine gets oxidized from 0 states to +5 states and reduced to -1 state respectively.

$${\text{ }}(0){\text{ }}( - 1){\text{ }}( + 5)( - 2) \\\ C{l_2} \to Cl + Cl{O_3}^ - \\\$$

In the reaction, we can see that the oxidation state of $Cl$ in $Cl{O_{3}}^ -$ is +5 oxidation states. $Cl$ loses five electrons to get oxidized and acts as a reducing agent. And also the oxidation state of Cl in Cl− is −1. Hence, Cl is reduced by the gaining of electrons.

Oxidation process -

$${\text{ }}0{\text{ }} + 5{\text{ }} - 3 \\\ C{l_2} \to 2Cl{O_3} + 10{e^ - } \\\$$

Reduction process -

$$0{\text{ }} - 1 \\\ Cl + 2{e^ - } \to 2C{l^ - }_{_{_{}}} \\\$$

(D) In the final reaction, we can understand that the $N{H_4}N{O_2}$ gets reduced to release nitrogen gas and water molecules. Therefore, both nitrogen and in $N{H_4}N{O_2}$ gets reduced from the (+5) and (+3) states to (0) states. Hence this is not an intramolecular redox reaction.

$${\text{ }}( + 5){\text{ }}( + 3){\text{ }}(0) \\\ N{H_4}N{O_2} \to {N_2} + 2{H_2}0 \\\$$

Note: We must understand the difference between the intermolecular redox reactions and an intramolecular redox reaction. Therefore Intermolecular means redox occurs between two molecules but in the case of an intramolecular reaction the redox process occurs within one molecule.