In acidic medium, H2O2 changes \text{Cr}_{2}\text{O}_{7}^{-2} to CrO5 which has two (-O-O-) bonds. Oxidation state of Cr in CrO5 is:

  • Option 1)

    +5

  • Option 2)

    +3

  • Option 3)

    +6

  • Option 4)

    -10

 

Answers (1)

As we learnt in

Rules for Oxidation Number -

The oxidation number of oxygen in most compounds is -2. There are two exceptions.
In peroxide (H2O2; Na2O2),  the oxidation number is -1.

- wherein

In superoxide e.g. KO2 its oxidation number is -1/2.

 

 CrO5 has butterfly structure having two peroxo bonds.

peroxo oxygen has - 1 oxidation state. Let oxidation state of Cr be x. 

CrO5 : x + 4( - 1) + 1 (-2) = 0 

\Rightarrow x = +6

Correct option is 3.

 


Option 1)

+5

Incorrect

Option 2)

+3

Incorrect

Option 3)

+6

Correct

Option 4)

-10

Incorrect

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