K2Cr2O7 + H2SO4 + 4H2O2 → K2SO4 + X + 5H2OWhat is the oxidation state of chromium in product ‘X’?

The reaction of K2Cr2O7, H2SO4, and H2O2 form K2SO4, CrO5, and H2O.K2Cr2O7+H2SO4+4H2O2→2CrO5(X)+K2SO4+5H2OStructure 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.

K2Cr2O7 + H2SO4 + 4H2O2 → K2SO4 + X + 5H2OWhat is the oxidation state of chromium in product ‘X’?

The reaction of K2Cr2O7, H2SO4, and H2O2 form K2SO4, CrO5, and H2O.K2Cr2O7+H2SO4+4H2O2→2CrO5(X)+K2SO4+5H2OStructure 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.

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$K_{2} {Cr}_{2} O_{7} + H_{2} {SO}_{4} + 4 H_{2} O_{2} \to K_{2} {SO}_{4} + X + 5 H_{2} O$
What is the oxidation state of chromium in product ‘X’?

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$+ 6$

$+ 10$

$+ 5$

$+ 7$

answer is B.

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Detailed Solution

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

$K_{2} {Cr}_{2} O_{7} + H_{2} {SO}_{4} + 4 H_{2} O_{2} \to \underset{(X)}{2 {CrO}_{5}} + K_{2} {SO}_{4} + 5 H_{2} O$

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.

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