In my book the oxidizing power in descending order of the following is given as: $\ce{BrO4-} > \ce{IO4-} > \ce{ClO4-}$
My doubt is regarding their order of oxidizing tendency.
I thought that chlorine has the highest electronegativity, so can help oxygen stabilize the $-1$ charge the best and be least reducing (electron donor) and most oxidizing. Looks like that’s not the case.
Another possibility was that iodine was the largest ion there, it’s far bigger than oxygen, so maybe it produced some instability which made the negative charge unstable, or in other words it became more reducing. In that case it should be least oxidizing (which is also not the case unfortunately).
I couldn't just make out whats so special about $\ce{BrO4-}$ that makes it the most oxidizing ion
Although this question gives a bit of a good hint but doesn't answer my question completely.