In acidic medium, H2O2 changes Cr2O72- to CrO5 which has two (-O-O-) bonds. Oxidation state of Cr in CrO5 is:

In acidic medium, H2O2 changes Cr2O72- to CrO5 which has two (-O-O-) bonds. Let the oxidation state of Cr in CrO5 be x.
The oxidation state of oxygen is -2.
Therefore, x + 5(-2) = 0
x - 10 = 0
x = +10
However, this is incorrect. The peroxo group (-O-O-) has an oxidation state of -2.
Therefore, the oxidation state of Cr in CrO5 is calculated as follows:
Let x be the oxidation state of Cr.
The oxidation state of oxygen in CrO5 is -2 except for the peroxo linkage (-O-O-) where each oxygen has an oxidation state of -1. Since there are two peroxo linkages, there are four oxygen atoms with an oxidation state of -1, and one oxygen atom with an oxidation state of -2.
Thus, x + 4(-1) + 1(-2) = 0
x - 4 - 2 = 0
x = +6
Therefore, the oxidation state of Cr in CrO5 is +6.