Idea of disproportionation

A species shows disproportionation if the same element in one oxidation state undergoes simultaneous oxidation and reduction to give two different oxidation states.

This is usually possible when the element is in an intermediate oxidation state. If it is already in its highest oxidation state, it cannot be oxidized further, so disproportionation is not possible.

Find oxidation state of Br in each species

• A: $BrO_4^-$

  Let oxidation state of Br be $x$. $x + 4(-2) = -1$ $x - 8 = -1$ $x = +7$

• B: $BrO^-$ $x + (-2) = -1$ $x = +1$

• C: $BrO_2^-$ $x + 2(-2) = -1$ $x - 4 = -1$ $x = +3$

• D: $BrO_3^-$ $x + 3(-2) = -1$ $x - 6 = -1$ $x = +5$

Check which oxidation state can disproportionate

For bromine, the oxidation states are: $-1, +1, +3, +5, +7$

• $+1$ can disproportionate to lower and higher states.
• $+3$ can disproportionate.
• $+5$ can disproportionate.
• $+7$ is the maximum oxidation state of bromine, so it cannot be oxidized further.

Therefore, $BrO_4^-$ does not show disproportionation.

Option check

• A: $BrO_4^-$ → Br is at $+7$ → does not disproportionate
• B: $BrO^-$ → Br is at $+1$ → can disproportionate
• C: $BrO_2^-$ → Br is at $+3$ → can disproportionate
• D: $BrO_3^-$ → Br is at $+5$ → can disproportionate

Final answer

The species that does not show disproportionation is: $\boxed{BrO_4^-}$ i.e. Option A.