The oxidation states of the central atoms in CrO ₅ and H ₂ SO ₅ are respectively:

Step-by-step solution

By standard rules, the oxidation state of Cr in CrO ₅ would be +10 ( x - 10 = 0 ) and S in H ₂ SO ₅ would be +8 ( 2 + x - 10 = 0 ). However, these exceed the maximum possible oxidation states for Cr (Group 6) and S (Group 16), which are both +6 . This discrepancy is due to the presence of peroxy linkages in their structures. In CrO ₅ (butterfly structure), there are two peroxy linkages (four oxygen atoms with -1 oxidation state) and one oxide linkage (one oxygen atom with -2 oxidation state). x + 4(-1) + 1(-2) = 0