Oxidising power of $C l O^{-}$ , $C l O_{2}^{-}$ , $C l O_{3}^{-}$ , $C l O_{4}^{-}$ is :

C l O_{4}^{-} > C l O_{2}^{-} > C l O_{3}^{-} > C l O^{-}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C l O_{2}^{-} > C l O^{-} > C l O_{3}^{-} > C l O_{4}^{-}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C l O^{-} < C l O_{2}^{-} < C l O_{3}^{-} < C l O_{4}^{-}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

C l O_{3}^{-} > C l O_{2}^{-} > C l O^{-} > C l O_{4}^{-}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $C l O^{-} < C l O_{2}^{-} < C l O_{3}^{-} < C l O_{4}^{-}$
Order of oxidation power: $C l O^{-} < C l O_{2}^{-} < C l O_{3}^{-} < C l O_{4}^{-}$ because the oxidation states of chloride in $C l O^{-}, C l O_{2}^{-}, C l O_{3}^{-}$ and $C l O_{4}^{-}$ are $+ 1, + 3, + 5$ and $+ 7$ respectively.
So, $C l O_{4}^{-}$ with highest oxidation state can easily oxidise and in turn get reduced.

Suggest Corrections

Similar questions

C l O^{-}, {C l O_{2}}^{-}, {C l O_{3}}^{-}, {C l O_{4}}^{-}

C l - O

C l O^{-}, C l O_{2}^{-}, C l O_{3}^{-}

C l O_{4}^{-}

C l O^{-}, C l O_{2}^{-}, C l O_{3}^{-}

C l O_{4}^{-}

C l O^{-}, C l O_{2}^{-}, C l O_{3}^{-}

C l O_{4}^{-}

3 C l O^{-} ⟶ C l O_{2}^{-} + 2 {C l}^{-}

C l O^{-} + C l O^{-} K_{1} - \to C l O_{2}^{-} + {C l}^{-}

C l O_{2}^{-} + C l O^{-} K_{2} - \to C l O_{3}^{-} + {C l}^{-}

Join BYJU'S Learning Program