${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7 + {\mathrm{H}}_2{\mathrm{SO}}_4 + 4{\mathrm{H}}_2{\mathrm{O}}_2 \to {\mathrm{K}}_2{\mathrm{SO}}_4 +\mathbf{ }\mathbf{X} + 5{\mathrm{H}}_2\mathrm{O}$

What is the oxidation state of chromium in product ‘X’?

The reaction of K₂Cr₂O₇, H₂SO₄, and H₂O₂ form K₂SO₄, CrO₅, and H₂O.

${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+{\mathrm{H}}_2{\mathrm{SO}}_4+4{\mathrm{H}}_2{\mathrm{O}}_2\to \underset{\left(\mathrm{X}\right)}{2{\mathrm{CrO}}_5}+{\mathrm{K}}_2{\mathrm{SO}}_4+5{\mathrm{H}}_2\mathrm{O}$

Structure of CrO₅ is:

In CrO₅, there are two peroxide linkages. The oxidation state of one oxygen atom of a peroxide linkage is -1. The oxidation state of Cr in CrO₅ is +6.