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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.

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  • $\begingroup$ I would love it if you can find a duplicate for me because anyways i am not getting an answer here @orthocresol $\endgroup$ – Shubham May 13 '16 at 8:19
  • $\begingroup$ Sorry Shubham, @orthocresol is right -- his answer to the stability question also answers the oxidizing power question. Bromine is the least stable of the three in the high $+7$ oxidation state... so what is it likely to do to become more stable? Oxidize something! $\endgroup$ – hBy2Py May 13 '16 at 11:05
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    $\begingroup$ The question does approach the matter from a different angle, though, and so it may be useful to have it open. $\endgroup$ – hBy2Py May 13 '16 at 11:06
  • $\begingroup$ @brian okay i take ur point that it may have some common points regarding bro4-. But what about the order of the other 2 ions? The question doesnt only ask about bro4- $\endgroup$ – Shubham May 13 '16 at 11:13
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    $\begingroup$ @orthocresol This is not a dupe, because it is not immediately apparant from the question that these will go in the same direction. For a dupe to be a dupe we need the essentially same answers and the essentially identical question. This one approaches the problem from a different angle. $\endgroup$ – Jan May 13 '16 at 11:18

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