Reduction potentials of some ions are given below. Arrange them in decreasing order of oxidising power.
$\begin{array}{llll} Ion {ClO}_{4}^{-} {IO}_{4}^{-} {BrO}_{4}^{-} \\ Reduction E^{Θ} = 1.19 V E^{Θ} = 1.65 V E^{Θ} = 1.74 V \\ potential E^{Θ} / V \end{array}$

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${IO}_{4}^{-} > {BrO}_{4}^{-} > {ClO}_{4}^{-}$

${ClO}_{4}^{-} > {IO}_{4}^{-} > {BrO}_{4}^{-}$

${BrO}_{4}^{-} > {ClO}_{4}^{-} > {IO}_{4}^{-}$

${BrO}_{4}^{-} > {IO}_{4}^{-} > {ClO}_{4}^{-}$

answer is A.

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Detailed Solution

Higher the reduction potential, higher is its tendency to get reduced. Hence, the order of oxidising power is:

${ClO}_{4}^{-} > {IO}_{4}^{-} > {BrO}_{4}^{-}$

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