Question

Which is the strongest oxidizing agent among $ClO_{4}^{-},BrO_{4}^{-}$ and $IO_{4}^{-}$ ?

Answer 1

Hint: Oxidizing agent is a chemical that will accept electrons and reduce itself and oxidize the other chemicals. Reducing agent is a chemical that will donate the electrons and oxidizes itself and reduces the other chemicals.

Complete answer:
- In the given question it is asked to find the strong oxidizing agent among the given options.
- In the question there are three chemicals $ClO_{4}^{-},BrO_{4}^{-}$ and $IO_{4}^{-}$ .
- In all the given chemicals there is a presence of $p\pi -d\pi $ back bonding.
- In $ClO_{4}^{-}$ , there is $2p\pi -3d\pi $bonding in between chlorine and oxygen.
- In $BrO_{4}^{-}$ , there is $2p\pi -4d\pi $bonding in between chlorine and oxygen.
- In $IO_{4}^{-}$ , there are $2p\pi -5d\pi $bonding in between chlorine and oxygen.
- Means the electrons in p-orbitals of the oxygen atom are donated to d-orbitals of the halogen atom.
- We know that the size of the d-orbitals is increasing as we are moving top to bottom in the periodic table then the overlapping will be less.
- Therefore in $ClO_{4}^{-}$ the overlapping is very effective when compared to $BrO_{4}^{-}$ and $IO_{4}^{-}$ .
- So the overlapping between the d-orbitals of a halogen atom with p-orbital of oxygen will be very less in $IO_{4}^{-}$ because of the larger size of the d- orbital in iodine.
- Therefore $IO_{4}^{-}$ will get reduced easily means will accept electrons from others very easily.
- So, $IO_{4}^{-}$ is a strong oxidizing agent among the given chemicals.

Note:
The size of the d-orbital in the oxidizing agent plays a big role to decide whether the chemical should be a strong oxidizing or weak oxidizing agent. If the d-orbital size is high it can accommodate more electrons.