Question

Which is the strongest oxidizing agent among $Cl{O_4}^{-}$ , $$Br{O_4}^{-}$$ and $$I{O_4}^{-}$$ ?

Answer 1

Hint:An oxidizing agent is one which itself gets reduced that means it actually has a tendency to accept electrons. Knowledge of p-dpi back bonding is must for the correct answer.

Complete step-by-step answer: An oxidizing agent is one which itself gets reduced that means it actually has a tendency to accept electrons.
 In among all the given ions there is p$\pi$−d$\pi$ back bonding.
 In ​ $Cl{O_4}^{-}$ , it is $\pi$−3d$\pi$, $$Br{O_4}^{-}$$ , it is 2p$\pi$−4d$\pi$ and in $$I{O_4}^{-}$$ , it is 2p$\pi$−5d$\pi$.
 Here few electrons from p orbitals of Oxygen are donated into the d−orbitals of atoms of halogens.
 As we can see the size of the d-orbital is increasing and thereby the effective overlapping will become less. Hence, $Cl{O_4}^{-}$ has the most effective overlapping which actually means the desire for electrons in Cl is well satisfied by the back bonding which is really less in the case of $$Br{O_4}^{-}$$ as here the overlapping is not so effective because of the large size of the d−orbitals. Therefore, Br can still accept electrons, which means it can be reduced more easily than the Cl and hence it acts as a strongest oxidizing agent among Cl and Br. And also I can get more reduced than Br.

Hence, order is:
$$C{l_4}^{-}<Br{O_4}^{-}<I{O_4}^{-}$$

Additional Information:
Oxidation is the process of losing electrons, as well as gaining oxygen or losing hydrogen. Oxidizing agent is the substance which is used to donate electrons or oxygen or gain hydrogen. In simpler words, the oxidizing agent itself gets reduced.
As it is clearly in the periodic table of elements, the halogens which are good oxidizing agents generally are fluorine, chlorine, bromine and iodine, and with fluorine is being the strongest oxidizing agent among the all, followed by chlorine, bromine and iodine. Another oxidizing agent is also oxygen.

Note:Among the all three, $$I{O_4}^{-}$$ is the strongest oxidising agent as it actually has a higher tendency of accepting electrons. . In among all the given ions there is p$\pi$−d$\pi$ back bonding and in the correct answer, it is 2p$\pi$−5d$\pi$.