$3 N_{2} H_{4} + 2 {B r O}_{3}^{-} ⟶ 3 N_{2} + 2 B r^{-} + 6 H_{2} O$

The above reaction shows :

intermolecular redox

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auto-redox

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intramolecular redox

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either of these

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Solution

The correct option is A intermolecular redox
The balanced reaction is as follows:

$3 N_{2} H_{4} + 2 {B r O}_{3}^{-} ⟶ 3 N_{2} + 2 B r^{-} + 6 H_{2} O$
The oxidation number of N increases from -2 to 0. Thus, $N_{2} H_{4}$ is oxidized.
The oxidation number of Br decreases from +5 to -1. Thus, ${B r O_{3}}^{-}$ is reduced.

Here, one compound oxidises while other reduces.Therefore, it is an intermolecular redox process.

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3 N_{2} H_{4} + 2 B r O_{3}^{-} \to 3 N_{2} + 2 B r^{-} + 6 H_{2} O

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3 N_{2} H_{4} + 2 {B r O}_{3}^{-} ⟶ 3 N_{2} + 2 B r^{-} + 6 H_{2} O

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