Question

In the reaction equation BrO-3(aq) Br-(aq) + BrO-4(aq), how many oxidation states does the disproportionate substance have throughout the reaction? 0 1 2 3

Answer 1

3

Step-by-step explanation:

Answer 2

Answer : The correct answer is 3.

Explanation :

The given chemical reaction is,

`BrO^-_3(aq)\rightarrow Br^-(aq)+BrO^-_4(aq)`

In a disproportionate reaction, disproportionate substance is a single substance which is oxidized as well as reduced.

The oxidation state of Br in
`BrO^-_3` is (+5).

The oxidation state of Br in
`Br^-` is (-1).

The oxidation state of Br in
`BrO^-_4` is (+7).

In the given chemical reaction,
`BrO^-_3` is getting reduced to
`Br^-` and
`BrO^-_3` is getting oxidized to
`BrO^-_4`. So, bromine is a disproportionate substance.

That means the bromine is present in three oxidation states.

Therefore, three oxidation states of the disproportionate substance have throughout the reaction.