Number of oxygen atoms in peroxide linkage in the product formed by reaction of $K_{2} C r_{2} O_{7}$ with $H_{2} O_{2}$ in acidic medium is

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Solution

The reaction is
$C r_{2} O_{7}^{2 -} + 4 H_{2} O_{2} + 2 H^{+} \to 2 C r O_{5} + 5 H_{2} O$
In $C r O_{5}$ there are two peroxide linkages, so number of oxygen atoms in peroxide linkage are 4.

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Number of oxygen atoms in peroxide linkage in the product formed by reaction of K2Cr2O7 with H2O2 in acidic medium is

The reaction is Cr2O2−7+4H2O2+2H+→2CrO5+5H2O In CrO5 there are two peroxide linkages, so number of oxygen atoms in peroxide linkage are 4.

Number of oxygen atoms in peroxide linkage in the product formed by reaction of $K_{2} C r_{2} O_{7}$ with $H_{2} O_{2}$ in acidic medium is

The reaction is
$C r_{2} O_{7}^{2 -} + 4 H_{2} O_{2} + 2 H^{+} \to 2 C r O_{5} + 5 H_{2} O$
In $C r O_{5}$ there are two peroxide linkages, so number of oxygen atoms in peroxide linkage are 4.