Animated Solution for Chemistry - Redox Reactions: The species given below that does not show disproportionation reaction is

Imagine a chemical species acting as both the hero and the villain in its own story. This is the essence of a disproportionation reaction. It is a fascinating type of redox reaction where a single element in a specific oxidation state simultaneously undergoes both oxidation (losing electrons to reach a higher state) and reduction (gaining electrons to reach a lower state).

For an element to pull off this dual role, there is a strict mathematical and physical constraint: its current oxidation state must lie strictly between its absolute minimum and absolute maximum possible oxidation states.

Think of it like standing on a staircase. If you are on the middle steps, you can choose to go up or go down. But if you are on the very top step, you can only go down. If you are on the bottom step, you can only go up.

Let's apply this logic to the element in our question: Bromine ($extBr$). Bromine is a proud member of Group 17, the halogens. It has 7 valence electrons.

To achieve a stable noble gas configuration, it can gain 1 electron, giving it a minimum oxidation state of $-1$.

Conversely, if it shares all 7 of its valence electrons with highly electronegative atoms (like oxygen), it can reach a maximum oxidation state of $+7$.

Therefore, for any bromine-containing species to undergo disproportionation, the oxidation state of bromine must be strictly greater than $-1$ and strictly less than $+7$.

Let's calculate the oxidation state of bromine in each of the given oxoanions. We know that oxygen generally exhibits an oxidation state of $-2$.

Option (b): $ext{BrO}^{-}$ (Hypobromite ion) Let the oxidation state of $extBr$ be $x$. Since $+1$ is between $-1$ and $+7$, $ext{BrO}^{-}$ can disproportionate.

Option (c): $ext{BrO}_2^{-}$ (Bromite ion) Since $+3$ is between $-1$ and $+7$, $ext{BrO}_2^{-}$ can disproportionate.

Option (d): $ext{BrO}_3^{-}$ (Bromate ion) Since $+5$ is between $-1$ and $+7$, $ext{BrO}_3^{-}$ can disproportionate.

Perbromate

Now, let's look at Option (a): $ext{BrO}_4^{-}$ (Perbromate ion).

Here is the catch! In the perbromate ion, bromine is sitting exactly at its maximum possible oxidation state of $+7$. It has already 'lost' or shared all its valence electrons. It is physically impossible for it to be oxidized any further to a higher state like $+8$.

Because it cannot be oxidized, it cannot fulfill the dual requirement of a disproportionation reaction. It can only act as an oxidizing agent (getting reduced itself). Therefore, $ext{BrO}_4^{-}$ is the species that does not show a disproportionation reaction.